NCERT Solutions for Class 11 Maths Chapter 4 - Principle Of Mathematical Induction

Exercise 4.1: This exercise consists of 12 questions, which are all based on proving statements using the principle of mathematical induction. These questions are designed to help students understand how to apply the principle of mathematical induction to prove various statements.

The questions in this exercise require students to prove various statements, such as:

• The sum of the first n odd natural numbers is n^2.

• The sum of the cubes of the first n natural numbers is [(n(n+1))/2]^2.

• The sum of the squares of the first n natural numbers is [n(n+1)(2n+1)]/6.

• In each question, students are required to first prove the statement for the base case (n = 1) and then assume the statement to be true for an arbitrary value of n (known as the induction hypothesis) before proving that the statement holds true for n+1.

The solutions to all the questions in Exercise 4.1 are provided in the NCERT Solutions for Class 11 Maths Chapter 4, which can help students check their answers and understand the concepts better.

Access NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction

1. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1+3+}{{\text{3}}^{\text{2}}}\text{+}.....\text{+}{{\text{3}}^{\text{n-1}}}\text{=}\dfrac{\left( {{\text{3}}^{\text{n}}}\text{-1} \right)}{\text{2}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1+3+}{{\text{3}}^{\text{2}}}\text{+}.....\text{+}{{\text{3}}^{\text{n-1}}}\text{=}\dfrac{\left( {{\text{3}}^{\text{n}}}\text{-1} \right)}{\text{2}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.}={{\text{3}}^{1-1}}\text{=1}\]

$\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\left( {{\text{3}}^{\text{1}}}\text{-1} \right)}{\text{2}}\text{=}\dfrac{\text{3-1}}{\text{2}}\text{=}\dfrac{\text{2}}{\text{2}}\text{=1}$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1+3+}{{\text{3}}^{\text{2}}}\text{+}.....\text{+}{{\text{3}}^{\text{k-1}}}\text{=}\dfrac{\left( {{\text{3}}^{\text{k}}}\text{-1} \right)}{\text{2}}\,\,.....\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1+3+}{{\text{3}}^{\text{2}}}\text{+}.....\text{+}{{\text{3}}^{\text{k-1}}}\text{+}{{\text{3}}^{\left( \text{k+1} \right)\text{-1}}}$ 

$\text{=}\left( \text{1+3+}{{\text{3}}^{\text{2}}}\text{+}....\text{+}{{\text{3}}^{\text{k-1}}} \right)\text{+}{{\text{3}}^{\text{k}}}$ 

$\text{=}\dfrac{\left( {{\text{3}}^{\text{k}}}\text{-1} \right)}{\text{2}}\text{+}{{\text{3}}^{\text{k}}}$          [Using (i)]

$\text{=}\dfrac{\left( {{\text{3}}^{\text{k}}}\text{-1} \right)\text{+2}\text{.}{{\text{3}}^{\text{k}}}}{\text{2}}$ 

$=\dfrac{\left( 1+2 \right){{3}^{k}}-1}{2}$ 

$=\dfrac{{{3.3}^{k}}-1}{2}$ 

$=\dfrac{{{3}^{k+1}}-1}{2}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

2. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N:}$ 

${{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{+}.....\text{+}{{\text{n}}^{\text{3}}}\text{=}{{\left( \dfrac{\text{n}\left( \text{n+1} \right)}{\text{2}} \right)}^{\text{2}}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

\[\text{P}\left( \text{1} \right)\text{:}{{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{+}.....\text{+}{{\text{n}}^{\text{3}}}\text{=}{{\left( \dfrac{\text{n}\left( \text{n+1} \right)}{\text{2}} \right)}^{\text{2}}}\] 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.}={{1}^{3}}\text{=1}\]

$\text{R}\text{.H}\text{.S}\text{.=}{{\left( \dfrac{1\left( \text{1+1} \right)}{\text{2}} \right)}^{\text{2}}}={{\left( 1 \right)}^{\text{2}}}=1$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

\[{{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{+}.....\text{+}{{\text{k}}^{\text{3}}}\text{=}{{\left( \dfrac{\text{k}\left( \text{k+1} \right)}{\text{2}} \right)}^{\text{2}}}\,\,.....\text{(i)}\] 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

\[{{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{+}.....\text{+}{{\text{k}}^{\text{3}}}\text{+}{{\left( \text{k+1} \right)}^{\text{3}}}\] 

$\text{=}\left( {{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{+}.....\text{+}{{\text{k}}^{\text{3}}} \right)\text{+}{{\left( \text{k+1} \right)}^{\text{3}}}$ 

$\text{=}{{\left( \dfrac{\text{k}\left( \text{k+1} \right)}{\text{2}} \right)}^{\text{2}}}\text{+}{{\left( \text{k+1} \right)}^{\text{3}}}$              [Using (i)]

$\text{=}\dfrac{{{\text{k}}^{\text{2}}}{{\left( \text{k+1} \right)}^{\text{2}}}}{\text{4}}\text{+}{{\left( \text{k+1} \right)}^{\text{3}}}$ 

$\text{=}\dfrac{{{\text{k}}^{\text{2}}}{{\left( \text{k+1} \right)}^{\text{2}}}\text{+4}{{\left( \text{k+1} \right)}^{\text{3}}}}{\text{4}}$ 

$\text{=}\dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}\left\{ {{\text{k}}^{\text{2}}}\text{+4}\left( \text{k+1} \right) \right\}}{\text{4}}$ 

$\text{=}\dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}\left\{ {{\text{k}}^{\text{2}}}\text{+4k+4} \right\}}{\text{4}}$ 

$\text{=}\dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}{{\left( \text{k+2} \right)}^{\text{2}}}}{\text{4}}$ 

$\text{=}\dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}{{\left( \text{k+1+1} \right)}^{\text{2}}}}{\text{4}}$ 

$\text{=}{{\left( \dfrac{\left( \text{k+1} \right)\left( \text{k+1+1} \right)}{\text{2}} \right)}^{\text{2}}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

3. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$ :

$\text{1+}\dfrac{\text{1}}{\left( \text{1+2} \right)}\text{+}\dfrac{\text{1}}{\left( \text{1+2+3} \right)}\text{+}.....\text{+}\dfrac{\text{1}}{\left( \text{1+2+3+}...\text{n} \right)}\text{=}\dfrac{\text{2n}}{\left( \text{n+1} \right)}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1+}\dfrac{\text{1}}{\text{1+2}}\text{+}\dfrac{\text{1}}{\text{1+2+3}}\text{+}.....\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{n}}\text{=}\dfrac{\text{2n}}{\text{n+1}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{1}{1}=1\]

$\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{2}\cdot \text{1}}{\text{1+1}}=1$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1+}\dfrac{\text{1}}{\left( \text{1+2} \right)}\text{+}\dfrac{\text{1}}{\left( \text{1+2+3} \right)}\text{+}.....\text{+}\dfrac{\text{1}}{\left( \text{1+2+3+}...\text{+k} \right)}\text{=}\dfrac{\text{2k}}{\left( \text{k+1} \right)}......(i)$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1+}\dfrac{\text{1}}{\text{1+2}}\text{+}\dfrac{\text{1}}{\text{1+2+3}}\text{+}....\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{+k}}\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{+k+}\left( \text{k+1} \right)}$ 

$\text{=}\left( \text{1+}\dfrac{\text{1}}{\text{1+2}}\text{+}\dfrac{\text{1}}{\text{1+2+3}}\text{+}...\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{+k}} \right)\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{+k+}\left( \text{k+1} \right)}$ 

$\text{=}\dfrac{\text{2k}}{\text{k+1}}\text{+}\dfrac{\text{1}}{\text{1+2+3+}...\text{+k+}\left( \text{k+1} \right)}$            [Using (i)]

$\text{=}\dfrac{\text{2k}}{\text{k+1}}\text{+}\dfrac{\text{1}}{\left( \dfrac{\left( \text{k+1} \right)\left( \text{k+1+1} \right)}{\text{2}} \right)}$            $\left[ \text{1+2+3}...\text{+n=}\dfrac{\text{n}\left( \text{n+1} \right)}{\text{2}} \right]$ 

$\text{=}\dfrac{\text{2k}}{\text{k+1}}\text{+}\dfrac{\text{2}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}$ 

$\text{=}\dfrac{\text{2}}{\left( \text{k+1} \right)}\left( \text{k+}\dfrac{\text{1}}{\text{k+2}} \right)$ 

$\text{=}\dfrac{\text{2}}{\left( \text{k+1} \right)}\left( \dfrac{{{\text{k}}^{\text{2}}}\text{+2k+1}}{\text{k+2}} \right)$ 

$\text{=}\dfrac{\text{2}}{\left( \text{k+1} \right)}\left[ \dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}}{\text{k+2}} \right]$ 

$\text{=}\dfrac{\text{2}\left( \text{k+1} \right)}{\left( \text{k+2} \right)}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

4. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1}\text{.2}\text{.3+2}\text{.3}\text{.4+}...\text{+n}\left( \text{n+1} \right)\left( \text{n+2} \right)\text{=}\dfrac{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)\left( \text{n+3} \right)}{\text{4}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1}\text{.2}\text{.3+2}\text{.3}\text{.4+}...\text{+n}\left( \text{n+1} \right)\left( \text{n+2} \right)\text{=}\dfrac{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)\left( \text{n+3} \right)}{\text{4}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=1}\text{.2}\text{.3=6}\]

$\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}\left( \text{1+1} \right)\left( \text{1+2} \right)\left( \text{1+3} \right)}{\text{4}}\text{=}\dfrac{\text{1}\text{.2}\text{.3}\text{.4}}{\text{4}}\text{=6}$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1}\text{.2}\text{.3+2}\text{.3}\text{.4+}...\text{+k}\left( \text{k+1} \right)\left( \text{k+2} \right)\text{=}\dfrac{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}{\text{4}}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1}\text{.2}\text{.3+2}\text{.3}\text{.4+}...\text{+k}\left( \text{k+1} \right)\left( \text{k+2} \right)\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)$ 

$\text{=}\left\{ \text{1}\text{.2}\text{.3+2}\text{.3}\text{.4+}...\text{+k}\left( \text{k+1} \right)\left( \text{k+2} \right) \right\}\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)\text{+}\left( \text{k+3} \right)$

$\text{=}\dfrac{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}{\text{4}}\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)$     [Using (i)]

$\text{=}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)\left( \dfrac{\text{k}}{\text{4}}\text{+1} \right)$

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)\left( \text{k+4} \right)}{\text{4}}$

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{k+1+1} \right)\left( \text{k+1+2} \right)\left( \text{k+1+3} \right)}{\text{4}}$

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

5. Prove the following by using the principle of mathematical induction for all \[\text{n}\in \text{N}\]:

$\text{1}\text{.3+2}\text{.}{{\text{3}}^{\text{2}}}\text{+3}\text{.}{{\text{3}}^{\text{3}}}\text{+}...\text{+n}\text{.}{{\text{3}}^{\text{n}}}\text{=}\dfrac{\left( \text{2n-1} \right){{\text{3}}^{\text{n+1}}}\text{+3}}{\text{4}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1}\text{.3+2}\text{.}{{\text{3}}^{\text{2}}}\text{+3}\text{.}{{\text{3}}^{\text{3}}}\text{+}...\text{+n}{{\text{3}}^{\text{n}}}\text{=}\dfrac{\left( \text{2n-1} \right){{\text{3}}^{\text{n+1}}}\text{+3}}{\text{4}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=1}\text{.}{{\text{3}}^{1}}=3\]

$\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\left( \text{2}\text{.1-1} \right){{\text{3}}^{\text{1+1}}}\text{+3}}{\text{4}}\text{=}\dfrac{{{\text{3}}^{\text{2}}}\text{+3}}{\text{4}}\text{=}\dfrac{\text{12}}{\text{4}}\text{=3}$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1}\text{.3+2}\text{.}{{\text{3}}^{\text{2}}}\text{+3}\text{.}{{\text{3}}^{\text{3}}}\text{+}...\text{+k}{{\text{3}}^{\text{k}}}\text{=}\dfrac{\left( \text{2k-1} \right){{\text{3}}^{\text{k+1}}}\text{+3}}{\text{4}}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1}\text{.3+2}\text{.}{{\text{3}}^{\text{2}}}\text{+3}\text{.}{{\text{3}}^{\text{3}}}\text{+}...\text{+k}\text{.}{{\text{3}}^{\text{k}}}\text{+}\left( \text{k+1} \right)\text{.}{{\text{3}}^{\text{k+1}}}$ 

$\text{=}\left( \text{1}\text{.3+2}\text{.}{{\text{3}}^{\text{2}}}\text{+3}\text{.}{{\text{3}}^{\text{3}}}\text{+}...\text{+k}\text{.}{{\text{3}}^{\text{k}}} \right)\text{+}\left( \text{k+1} \right)\text{.}{{\text{3}}^{\text{k+1}}}$ 

$\text{=}\dfrac{\left( \text{2k-1} \right){{\text{3}}^{\text{k+1}}}\text{+3}}{\text{4}}\text{+}\left( \text{k+1} \right){{\text{3}}^{\text{k-1}}}$           [Using (i)]

$\text{=}\dfrac{\left( \text{2k-1} \right){{\text{3}}^{\text{k+1}}}\text{+3+4}\left( \text{k+1} \right){{\text{3}}^{\text{k+1}}}}{\text{4}}$ 

$\text{=}\dfrac{{{\text{3}}^{\text{k+1}}}\left\{ \text{2k-1+4}\left( \text{k+1} \right) \right\}\text{+3}}{\text{4}}$ 

$\text{=}\dfrac{{{\text{3}}^{\text{k+1}}}\left\{ \text{6k+3} \right\}\text{+3}}{\text{4}}$ 

$\text{=}\dfrac{{{\text{3}}^{\text{k+1}}}\text{.3}\left\{ \text{2k+1} \right\}\text{+3}}{\text{4}}$
$\text{=}\dfrac{{{\text{3}}^{\left( \text{k+1} \right)\text{+1}}}\left\{ \text{2k+1} \right\}\text{+3}}{\text{4}}$ 

$\text{=}\dfrac{\left\{ \text{2}\left( \text{k+1} \right)\text{-1} \right\}{{\text{3}}^{\left( \text{k+1} \right)\text{+1}}}\text{+3}}{\text{4}}$ 

      Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

6. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1}\text{.2+2}\text{.3+3}\text{.4+}...\text{+n}\text{.}\left( \text{n+1} \right)\text{=}\left[ \dfrac{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)}{\text{3}} \right]$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1}\text{.2+2}\text{.3+3}\text{.4+}...\text{+n}\text{.}\left( \text{n+1} \right)\text{=}\left[ \dfrac{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)}{\text{3}} \right]$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=1}\text{.2=2}\]

$\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}\left( \text{1+1} \right)\left( \text{1+2} \right)}{\text{3}}\text{=}\dfrac{\text{1}\text{.2}\text{.3}}{\text{3}}\text{=2}$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1}\text{.2+2}\text{.3+3}\text{.4+}...\text{+k}\text{.}\left( \text{k+1} \right)\text{=}\left[ \dfrac{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)}{\text{3}} \right]...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1}\text{.2+2}\text{.3+3}\text{.4+}...\text{+k}\text{.}\left( \text{k+1} \right)\text{+}\left( \text{k+1} \right)\text{.}\left( \text{k+2} \right)$ 

$\text{=}\left[ \text{1}\text{.2+2}\text{.3+3}\text{.4+}...\text{+k}\text{.}\left( \text{k+1} \right) \right]\text{+}\left( \text{k+1} \right)\text{.}\left( \text{k+2} \right)$ 

$\text{=}\dfrac{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)}{\text{3}}\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)$         [Using (i)]

$\text{=}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \dfrac{\text{k}}{\text{3}}\text{+1} \right)$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{k+1+1} \right)\left( \text{k+1+2} \right)}{\text{3}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

7. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1}\text{.3+3}\text{.5+5}\text{.7+}...\text{+}\left( \text{2n-1} \right)\left( \text{2n+1} \right)\text{=}\dfrac{\text{n}\left( \text{4}{{\text{n}}^{\text{2}}}\text{+6n-1} \right)}{\text{3}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1}\text{.3+3}\text{.5+5}\text{.7+}...\text{+}\left( \text{2n-1} \right)\left( \text{2n+1} \right)\text{=}\dfrac{\text{n}\left( \text{4}{{\text{n}}^{\text{2}}}\text{+6n-1} \right)}{\text{3}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=1}\text{.3=3}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}\left( \text{4}\text{.}{{\text{1}}^{\text{2}}}\text{+6}\text{.1-1} \right)}{\text{3}}\text{=}\dfrac{\text{4+6-1}}{\text{3}}\text{=}\dfrac{\text{9}}{\text{3}}\text{=3}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1}\text{.3+3}\text{.5+5}\text{.7+}...\text{+}\left( \text{2k-1} \right)\left( \text{2k+1} \right)\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+6k-1} \right)}{\text{3}}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left( \text{1}\text{.3+3}\text{.5+5}\text{.7+}...\text{+}\left( \text{2k-1} \right)\left( \text{2k+1} \right) \right)\text{+}\left\{ \left( \text{k+1} \right)\text{-1} \right\}\left\{ \text{2}\left( \text{k+1} \right)\text{+1} \right\}$ 

$\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+6k-1} \right)}{\text{3}}\text{+}\left( \text{2k+2-1} \right)\left( \text{2k+2+1} \right)$   [Using (i)]

$\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+6k-1} \right)}{\text{3}}\text{+}\left( \text{2k+1} \right)\left( \text{2k+3} \right)$ 

$\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+6k-1} \right)}{\text{3}}\text{+}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+8k+3} \right)$ 

$\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+6k-1} \right)\text{+3}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+8k+3} \right)}{\text{3}}$ 

$\text{=}\dfrac{\text{4}{{\text{k}}^{\text{3}}}\text{+6}{{\text{k}}^{\text{2}}}\text{-k+12}{{\text{k}}^{\text{2}}}\text{+24k+9}}{\text{3}}$ 

$\text{=}\dfrac{\text{4}{{\text{k}}^{\text{3}}}\text{+18}{{\text{k}}^{\text{2}}}\text{+23k+9}}{\text{3}}$ 

$\text{=}\dfrac{\text{4}{{\text{k}}^{\text{3}}}\text{+14}{{\text{k}}^{\text{2}}}\text{+9k+4}{{\text{k}}^{\text{2}}}\text{+14k+9}}{\text{3}}$ 

$\text{=}\dfrac{\text{k}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+14k+9} \right)\text{+1}\left( \text{4}{{\text{k}}^{\text{2}}}\text{+14k+9} \right)}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{4}{{\text{k}}^{\text{2}}}\text{+14k+9} \right)}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left\{ \text{4}{{\text{k}}^{\text{2}}}\text{+8k+4+6k+6-1} \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left\{ \text{4}\left( {{\text{k}}^{\text{2}}}\text{+2k+1} \right)\text{+6}\left( \text{k+1} \right)\text{-1} \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left\{ \text{4}{{\left( \text{k+1} \right)}^{\text{2}}}\text{+6}\left( \text{k+1} \right)\text{-1} \right\}}{\text{3}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

8. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1}\text{.2+2}\text{.}{{\text{2}}^{\text{2}}}\text{+3}\text{.}{{\text{2}}^{\text{2}}}\text{+}....\text{+n}\text{.}{{\text{2}}^{\text{n}}}\text{=}\left( \text{n-1} \right){{\text{2}}^{\text{n+1}}}\text{+2}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1}\text{.2+2}\text{.}{{\text{2}}^{\text{2}}}\text{+3}\text{.}{{\text{2}}^{\text{2}}}\text{+}....\text{+n}\text{.}{{\text{2}}^{\text{n}}}\text{=}\left( \text{n-1} \right){{\text{2}}^{\text{n+1}}}\text{+2}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=1}\text{.2=2}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\left( \text{1-1} \right){{\text{2}}^{\text{1+1}}}\text{+2=0+2=2}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1}\text{.2+2}\text{.}{{\text{2}}^{\text{2}}}\text{+3}\text{.}{{\text{2}}^{\text{2}}}\text{+}...\text{+k}\text{.}{{\text{2}}^{\text{k}}}\text{=}\left( \text{k-1} \right){{\text{2}}^{\text{k+1}}}\text{+2}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left\{ \text{1}\text{.2+2}\text{.}{{\text{2}}^{\text{2}}}\text{+3}\text{.}{{\text{2}}^{\text{2}}}\text{+}...\text{+k}\text{.}{{\text{2}}^{\text{k}}} \right\}\text{+}\left( \text{k+1} \right)\text{.}{{\text{2}}^{\text{k+1}}}$ 

$\text{=}\left( \text{k-1} \right){{\text{2}}^{\text{k+1}}}\text{+2+}\left( \text{k+1} \right){{\text{2}}^{\text{k+1}}}$ 

$\text{=}{{\text{2}}^{\text{k+1}}}\left\{ \left( \text{k-1} \right)\text{+}\left( \text{k+1} \right) \right\}\text{+2}$ 

$\text{=}{{\text{2}}^{\text{k+1}}}\text{.2k+2}$ 

$\text{=k}\text{.}{{\text{2}}^{\left( \text{k+1} \right)\text{+1}}}\text{+2}$ 

$\text{=}\left\{ \left( \text{k+1} \right)\text{-1} \right\}{{\text{2}}^{\left( \text{k+1} \right)\text{+1}}}\text{+2}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

9. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

\[\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{8}}\text{+}...\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{n}}}}\text{=1-}\dfrac{\text{1}}{{{\text{2}}^{\text{n}}}}\] 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{8}}\text{+}...\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{n}}}}\text{=1-}\dfrac{\text{1}}{{{\text{2}}^{\text{n}}}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{1}{2}\]

\[\text{R}\text{.H}\text{.S}\text{.=1-}\dfrac{\text{1}}{{{\text{2}}^{\text{1}}}}\text{=}\dfrac{\text{1}}{\text{2}}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

\[\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{8}}\text{+}...\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}}\text{=1-}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}}...\text{(i)}\] 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left( \dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{4}}\text{+}\dfrac{\text{1}}{\text{8}}\text{+}...\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}} \right)\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{k+1}}}}$ 

$\text{=}\left( \text{1-}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}} \right)\text{+}\dfrac{\text{1}}{{{\text{2}}^{\text{k+1}}}}$      [Using (i)]

$\text{1-}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}}\left( \text{1-}\dfrac{\text{1}}{\text{2}} \right)$ 

$\text{=1-}\dfrac{\text{1}}{{{\text{2}}^{\text{k}}}}\left( \dfrac{\text{1}}{\text{2}} \right)$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

10. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\dfrac{\text{1}}{\text{2}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.8}}\text{+}\dfrac{\text{1}}{\text{8}\text{.11}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3n-1} \right)\left( \text{3n+2} \right)}\text{=}\dfrac{\text{n}}{\left( \text{6n+4} \right)}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}\dfrac{\text{1}}{\text{2}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.8}}\text{+}\dfrac{\text{1}}{\text{8}\text{.11}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3n-1} \right)\left( \text{3n+2} \right)}\text{=}\dfrac{\text{n}}{\left( \text{6n+4} \right)}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{2}\text{.5}}\text{=}\dfrac{\text{1}}{\text{10}}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{6}\text{.1+4}}\text{=}\dfrac{\text{1}}{\text{10}}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\dfrac{\text{1}}{\text{2}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.8}}\text{+}\dfrac{\text{1}}{\text{8}\text{.11}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3k+1} \right)\left( \text{3k+2} \right)}\text{=}\dfrac{\text{k}}{\left( \text{6k+4} \right)}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\dfrac{\text{1}}{\text{2}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.8}}\text{+}\dfrac{\text{1}}{\text{8}\text{.11}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3k+1} \right)\left( \text{3k+2} \right)}\text{+}\dfrac{\text{1}}{\left\{ \text{3}\left( \text{k+1} \right)\text{-1} \right\}\left\{ \text{3}\left( \text{k+1} \right)\text{+2} \right\}}$ 

$\text{=}\dfrac{\text{k}}{\text{6k+4}}\text{+}\dfrac{\text{1}}{\left( \text{3k+3-1} \right)\left( \text{3k+3+2} \right)}$        [Using (i)]

$\text{=}\dfrac{\text{k}}{\text{6k+4}}\text{+}\dfrac{\text{1}}{\left( \text{3k+2} \right)\left( \text{3k+5} \right)}$ 

$\text{=}\dfrac{\text{k}}{\text{2}\left( \text{3k+2} \right)}\text{+}\dfrac{\text{1}}{\left( \text{3k+2} \right)\left( \text{3k+5} \right)}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{3k+2} \right)}\left( \dfrac{\text{k}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{3k+5}} \right)$ 

     \[\text{=}\dfrac{\text{1}}{\left( \text{3k+2} \right)}\left( \dfrac{\text{k}\left( \text{3k+5} \right)\text{+2}}{\text{2}\left( \text{3k+5} \right)} \right)\] 

    $\text{=}\dfrac{\text{1}}{\left( \text{3k+2} \right)}\left( \dfrac{\text{3}{{\text{k}}^{\text{2}}}\text{+5k+2}}{\text{2}\left( \text{3k+5} \right)} \right)$ 

    $\text{=}\dfrac{\text{1}}{\left( \text{3k+2} \right)}\left( \dfrac{\left( \text{3k+2} \right)\left( \text{k+1} \right)}{\text{2}\left( \text{3k+5} \right)} \right)$ 

     $\text{=}\dfrac{\left( \text{k+1} \right)}{\text{6k+10}}$ 

     $\text{=}\dfrac{\left( \text{k+1} \right)}{\text{6}\left( \text{k+1} \right)\text{+4}}$     

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle 

of mathematical induction.

11. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\dfrac{\text{1}}{\text{1}\text{.2}\text{.3}}\text{+}\dfrac{\text{1}}{\text{2}\text{.3}\text{.4}}\text{+}\dfrac{\text{1}}{\text{3}\text{.4}\text{.5}}\text{+}...\text{+}\dfrac{\text{1}}{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)}\text{=}\dfrac{\text{n}\left( \text{n+3} \right)}{\text{4}\left( \text{n+1} \right)\left( \text{n+2} \right)}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}\dfrac{\text{1}}{\text{1}\text{.2}\text{.3}}\text{+}\dfrac{\text{1}}{\text{2}\text{.3}\text{.4}}\text{+}\dfrac{\text{1}}{\text{3}\text{.4}\text{.5}}\text{+}...\text{+}\dfrac{\text{1}}{\text{n}\left( \text{n+1} \right)\left( \text{n+2} \right)}\text{=}\dfrac{\text{n}\left( \text{n+3} \right)}{\text{4}\left( \text{n+1} \right)\left( \text{n+2} \right)}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{1}\text{.2}\text{.3}}=\dfrac{1}{6}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}\text{.}\left( \text{1+3} \right)}{\text{4}\left( \text{1+1} \right)\left( \text{1+2} \right)}\text{=}\dfrac{\text{1}\text{.4}}{\text{4}\text{.2}\text{.3}}\text{=}\dfrac{\text{1}}{6}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\dfrac{\text{1}}{\text{1}\text{.2}\text{.3}}\text{+}\dfrac{\text{1}}{\text{2}\text{.3}\text{.4}}\text{+}\dfrac{\text{1}}{\text{3}\text{.4}\text{.5}}\text{+}...\text{+}\dfrac{\text{1}}{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)}\text{=}\dfrac{\text{k}\left( \text{k+3} \right)}{\text{4}\left( \text{k+1} \right)\left( \text{k+2} \right)}...(i)$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left[ \dfrac{\text{1}}{\text{1}\text{.2}\text{.3}}\text{+}\dfrac{\text{1}}{\text{2}\text{.3}\text{.4}}\text{+}\dfrac{\text{1}}{\text{3}\text{.4}\text{.5}}\text{+}...\text{+}\dfrac{\text{1}}{\text{k}\left( \text{k+1} \right)\left( \text{k+2} \right)} \right]\text{+}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}$ 

$\text{=}\dfrac{\text{k}\left( \text{k+3} \right)}{\text{4}\left( \text{k+1} \right)\left( \text{k+2} \right)}\text{+}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}$      [Using (i)]

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{\text{k}\left( \text{k+3} \right)}{\text{4}}\text{+}\dfrac{\text{1}}{\text{k+3}} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{\text{k}{{\left( \text{k+3} \right)}^{\text{2}}}\text{+4}}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{\text{k}\left( {{\text{k}}^{\text{2}}}\text{+6k+9} \right)\text{+4}}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{{{\text{k}}^{\text{3}}}\text{+6}{{\text{k}}^{\text{2}}}\text{+9k+4}}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{{{\text{k}}^{\text{3}}}\text{+2}{{\text{k}}^{\text{2}}}\text{+k+4}{{\text{k}}^{\text{2}}}\text{+8k+4}}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{\text{k}\left( {{\text{k}}^{\text{2}}}\text{+2k+1} \right)\text{+4}\left( {{\text{k}}^{\text{2}}}\text{+2k+1} \right)}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{k+1} \right)\left( \text{k+2} \right)}\left\{ \dfrac{\text{k}{{\left( \text{k+1} \right)}^{\text{2}}}\text{+4}{{\left( \text{k+1} \right)}^{\text{2}}}}{\text{4}\left( \text{k+3} \right)} \right\}$ 

$\text{=}\dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}\left( \text{k+4} \right)}{\text{4}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+3} \right)}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left\{ \left( \text{k+1} \right)\text{+3} \right\}}{\text{4}\left\{ \left( \text{k+1} \right)\text{+1} \right\}\left\{ \left( \text{k+1} \right)\text{+2} \right\}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

12. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{a+ar+a}{{\text{r}}^{\text{2}}}\text{+}...\text{+a}{{\text{r}}^{\text{n-1}}}\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{n}}}\text{-1} \right)}{\text{r-1}}$

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:a+ar+a}{{\text{r}}^{\text{2}}}\text{+}...\text{+a}{{\text{r}}^{\text{n-1}}}\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{n}}}\text{-1} \right)}{\text{r-1}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=a}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{a}\left( {{\text{r}}^{\text{1}}}\text{-1} \right)}{\left( \text{r-1} \right)}\text{=a}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{a+ar+a}{{\text{r}}^{\text{2}}}\text{+}...\text{+a}{{\text{r}}^{\text{k-1}}}\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{k}}}\text{-1} \right)}{\text{r-1}}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left\{ \text{a+ar+a}{{\text{r}}^{\text{2}}}\text{+}...\text{+a}{{\text{r}}^{\text{k-1}}} \right\}\text{+a}{{\text{r}}^{\left( \text{k+1} \right)\text{-1}}}$ 

$\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{k}}}\text{-1} \right)}{\text{r-1}}\text{+a}{{\text{r}}^{\text{k}}}$      [Using (i)]

$\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{k}}}\text{-1} \right)\text{+a}{{\text{r}}^{\text{k}}}\left( \text{r-1} \right)}{\text{r-1}}$ 

$\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{k}}}\text{-1} \right)\text{+a}{{\text{r}}^{\text{k+1}}}\text{-a}{{\text{r}}^{\text{k}}}}{\text{r-1}}$ 

$\text{=}\dfrac{\text{a}{{\text{r}}^{\text{k}}}\text{-a+a}{{\text{r}}^{\text{k+1}}}\text{-a}{{\text{r}}^{\text{k}}}}{\text{r-1}}$ 

$\text{=}\dfrac{\text{a}{{\text{r}}^{\text{k+1}}}\text{-a}}{\text{r-1}}$ 

$\text{=}\dfrac{\text{a}\left( {{\text{r}}^{\text{k+1}}}\text{-1} \right)}{\text{r-1}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

13. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\left( \text{1+}\dfrac{\text{3}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{5}}{\text{4}} \right)\left( \text{1+}\dfrac{\text{7}}{\text{9}} \right)...\left( \text{1+}\dfrac{\left( \text{2n+1} \right)}{{{\text{n}}^{\text{2}}}} \right)\text{=}{{\left( \text{n+1} \right)}^{\text{2}}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

\[\text{P}\left( \text{n} \right)\text{:}\left( \text{1+}\dfrac{\text{3}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{5}}{\text{4}} \right)\left( \text{1+}\dfrac{\text{7}}{\text{9}} \right)...\left( \text{1+}\dfrac{\left( \text{2n+1} \right)}{{{\text{n}}^{\text{2}}}} \right)\text{=}{{\left( \text{n+1} \right)}^{\text{2}}}\] 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\left( \text{1+}\dfrac{\text{3}}{\text{1}} \right)\text{=4}\]

\[\text{R}\text{.H}\text{.S}\text{.=}{{\left( \text{1+1} \right)}^{\text{2}}}\text{=}{{\text{2}}^{\text{2}}}\text{=4}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

\[\left( \text{1+}\dfrac{\text{3}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{5}}{\text{4}} \right)\left( \text{1+}\dfrac{\text{7}}{\text{9}} \right)...\left( \text{1+}\dfrac{\left( \text{2k+1} \right)}{{{\text{k}}^{\text{2}}}} \right)\text{=}{{\left( \text{k+1} \right)}^{\text{2}}}...(i)\] 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left[ \left( \text{1+}\dfrac{\text{3}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{5}}{\text{4}} \right)\left( \text{1+}\dfrac{\text{7}}{\text{9}} \right)...\left( \text{1+}\dfrac{\left( \text{2k+1} \right)}{{{\text{k}}^{\text{2}}}} \right) \right]\left\{ \text{1+}\dfrac{\left\{ \text{2}\left( \text{k+1} \right)\text{+1} \right\}}{{{\left( \text{k+1} \right)}^{\text{2}}}} \right\}$ 

$\text{=}{{\left( \text{k+1} \right)}^{\text{2}}}\left( \text{1+}\dfrac{\text{2}\left( \text{k+1} \right)\text{+1}}{{{\left( \text{k+1} \right)}^{\text{2}}}} \right)$         [Using (i)]

$\text{=}{{\left( \text{k+1} \right)}^{\text{2}}}\left[ \dfrac{{{\left( \text{k+1} \right)}^{\text{2}}}\text{+2}\left( \text{k+1} \right)\text{+1}}{{{\left( \text{k+1} \right)}^{\text{2}}}} \right]$ 

$={{\left( \text{k+1} \right)}^{\text{2}}}+2\left( k+1 \right)+1$ 

$\text{=}{{\left\{ \left( \text{k+1} \right)\text{+1} \right\}}^{\text{2}}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

14. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\left( \text{1+}\dfrac{\text{1}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{2}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{3}} \right)...\left( \text{1+}\dfrac{\text{1}}{\text{n}} \right)\text{=}\left( \text{n+1} \right)$

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}\left( \text{1+}\dfrac{\text{1}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{2}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{3}} \right)...\left( \text{1+}\dfrac{\text{1}}{\text{n}} \right)\text{=}\left( \text{n+1} \right)$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\left( \text{1+}\dfrac{\text{1}}{\text{1}} \right)\text{=2}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\left( \text{1+1} \right)=2\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{P}\left( \text{k} \right)\text{:}\left( \text{1+}\dfrac{\text{1}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{2}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{3}} \right)...\left( \text{1+}\dfrac{\text{1}}{\text{n}} \right)\text{=}\left( \text{k+1} \right)...(i)$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left[ \text{P}\left( \text{k} \right)\text{:}\left( \text{1+}\dfrac{\text{1}}{\text{1}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{2}} \right)\left( \text{1+}\dfrac{\text{1}}{\text{3}} \right)...\left( \text{1+}\dfrac{\text{1}}{\text{k}} \right) \right]\left( \text{1+}\dfrac{\text{1}}{\text{k+1}} \right)$ 

$\text{=}\left( \text{k+1} \right)\left( \text{1+}\dfrac{\text{1}}{\text{k+1}} \right)$           [Using (i)]

$\text{=}\left( \text{k+1} \right)\left[ \dfrac{\left( \text{k+1} \right)\text{+1}}{\left( \text{k+1} \right)} \right]$ 

$\text{=}\left( \text{k+1} \right)\text{+1}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

15. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

${{\text{1}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}...\text{+}{{\left( \text{2n-1} \right)}^{\text{2}}}\text{=}\dfrac{\text{n}\left( \text{2n-1} \right)\left( \text{2n+1} \right)}{\text{3}}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}{{\text{1}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}...\text{+}{{\left( \text{2n-1} \right)}^{\text{2}}}\text{=}\dfrac{\text{n}\left( \text{2n-1} \right)\left( \text{2n+1} \right)}{\text{3}}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}{{\text{1}}^{\text{2}}}\text{=1}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}\left( \text{2}\text{.1-1} \right)\left( \text{2}\text{.1+1} \right)}{\text{3}}\text{=}\dfrac{\text{1}\text{.1}\text{.3}}{\text{3}}\text{=1}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{P}\left( \text{k} \right)\text{:}{{\text{1}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}...\text{+}{{\left( \text{2n-1} \right)}^{\text{2}}}\text{=}\dfrac{\text{k}\left( \text{2k-1} \right)\left( \text{2k+1} \right)}{\text{3}}...(i)$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left\{ {{\text{1}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}...\text{+}{{\left( \text{2k-1} \right)}^{\text{2}}} \right\}\text{+}{{\left\{ \text{2}\left( \text{k+1} \right)\text{-1} \right\}}^{\text{2}}}$ 

$\text{=}\dfrac{\text{k}\left( \text{2k-1} \right)\left( \text{2k+1} \right)}{\text{3}}\text{+}{{\left( \text{2k+2-1} \right)}^{\text{2}}}$      [Using (i)]

$\text{=}\dfrac{\text{k}\left( \text{2k-1} \right)\left( \text{2k+1} \right)}{\text{3}}\text{+}{{\left( \text{2k+1} \right)}^{\text{2}}}$ 

$\text{=}\dfrac{\text{2}\left( \text{2k-1} \right)\left( \text{2k+1} \right)\text{+3}{{\left( \text{2k+1} \right)}^{\text{2}}}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left\{ \text{k}\left( \text{2k-1} \right)\text{+3}\left( \text{2k+1} \right) \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left\{ \text{2}{{\text{k}}^{\text{2}}}\text{-k+6k+3} \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left\{ \text{2}{{\text{k}}^{\text{2}}}\text{+5k+3} \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left\{ \text{2}{{\text{k}}^{\text{2}}}\text{+2k+3k+3} \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left\{ \text{2k}\left( \text{k+1} \right)\text{+3}\left( \text{k+1} \right) \right\}}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{2k+1} \right)\left( \text{k+1} \right)\left( \text{2k+3} \right)}{\text{3}}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left\{ \text{2}\left( \text{k+1} \right)\text{-1} \right\}\left\{ \text{2}\left( \text{k+1} \right)\text{+1} \right\}}{\text{3}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

16. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\dfrac{\text{1}}{\text{1}\text{.4}}\text{+}\dfrac{\text{1}}{\text{4}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.10}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3n-2} \right)\left( \text{3n+1} \right)}\text{=}\dfrac{\text{n}}{\left( \text{3n+1} \right)}$ 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:}\dfrac{\text{1}}{\text{1}\text{.4}}\text{+}\dfrac{\text{1}}{\text{4}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.10}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3n-2} \right)\left( \text{3n+1} \right)}\text{=}\dfrac{\text{n}}{\left( \text{3n+1} \right)}$ 

For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{1}{1.4}=\dfrac{1}{4}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{3}\text{.1+1}}\text{=}\dfrac{\text{1}}{\text{4}}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{P}\left( \text{k} \right)\text{:}\dfrac{\text{1}}{\text{1}\text{.4}}\text{+}\dfrac{\text{1}}{\text{4}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.10}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3k-2} \right)\left( \text{3k+1} \right)}\text{=}\dfrac{\text{k}}{\left( \text{3k+1} \right)}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left\{ \dfrac{\text{1}}{\text{1}\text{.4}}\text{+}\dfrac{\text{1}}{\text{4}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.10}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{3k-2} \right)\left( \text{3k+1} \right)} \right\}\text{+}\dfrac{\text{1}}{\left\{ \text{3}\left( \text{k+1} \right)\text{-2} \right\}\left\{ \text{3}\left( \text{k+1} \right)\text{+1} \right\}}$ 

$\text{=}\dfrac{\text{k}}{\text{3k+1}}\text{+}\dfrac{\text{1}}{\left( \text{3k+1} \right)\left( \text{3k+4} \right)}$        [Using (i)]

$\text{=}\dfrac{\text{1}}{\left( \text{3k+1} \right)}\left\{ \text{k+}\dfrac{\text{1}}{\left( \text{3k+4} \right)} \right\}$

$\text{=}\dfrac{\text{1}}{\left( \text{3k+1} \right)}\left\{ \dfrac{\text{k}\left( \text{3k+4} \right)\text{+1}}{\left( \text{3k+4} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{3k+1} \right)}\left\{ \dfrac{\text{3}{{\text{k}}^{\text{2}}}\text{+4k+1}}{\left( \text{3k+4} \right)} \right\}$ 

$\text{=}\dfrac{\text{1}}{\left( \text{3k+1} \right)}\left\{ \dfrac{\text{3}{{\text{k}}^{\text{2}}}\text{+3k+k+1}}{\left( \text{3k+4} \right)} \right\}$ 

$\text{=}\dfrac{\left( \text{3k+1} \right)\left( \text{k+1} \right)}{\left( \text{3k+1} \right)\left( \text{3k+4} \right)}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)}{\text{3}\left( \text{k+1} \right)\text{+1}}$  

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

17. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\dfrac{\text{1}}{\text{3}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.9}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{2n+1} \right)\left( \text{2n+3} \right)}\text{=}\dfrac{\text{n}}{\text{2}\left( \text{2n+3} \right)}$ 

Ans: For $\text{n=1}$, 

\[\text{L}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{3}\text{.5}}\]

\[\text{R}\text{.H}\text{.S}\text{.=}\dfrac{\text{1}}{\text{3}\left( \text{2}\text{.1+3} \right)}\text{=}\dfrac{\text{1}}{\text{3}\text{.5}}\]

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{P}\left( \text{k} \right)\text{:}\dfrac{\text{1}}{\text{3}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.9}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{2k+1} \right)\left( \text{2k+3} \right)}\text{=}\dfrac{\text{k}}{\text{2}\left( \text{2k+3} \right)}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\left[ \dfrac{\text{1}}{\text{3}\text{.5}}\text{+}\dfrac{\text{1}}{\text{5}\text{.7}}\text{+}\dfrac{\text{1}}{\text{7}\text{.9}}\text{+}...\text{+}\dfrac{\text{1}}{\left( \text{2k+1} \right)\left( \text{2k+3} \right)} \right]\text{+}\dfrac{\text{1}}{\left\{ \text{2}\left( \text{k+1} \right)\text{+1} \right\}\left\{ \text{2}\left( \text{k+1} \right)\text{+3} \right\}}$ 

$\text{=}\dfrac{\text{k}}{\text{3}\left( \text{2k+3} \right)}\text{+}\dfrac{\text{1}}{\left( \text{2k+3} \right)\left( \text{2k+5} \right)}$           [Using (i)]

$\text{=}\dfrac{\text{1}}{\left( \text{2k+3} \right)}\left[ \dfrac{\text{k}}{\text{3}}\text{+}\dfrac{\text{1}}{\left( \text{2k+5} \right)} \right]$ 

$\text{=}\dfrac{\text{1}}{\left( \text{2k+3} \right)}\left[ \dfrac{\text{k}\left( \text{2k+5} \right)\text{+3}}{\text{3}\left( \text{2k+5} \right)} \right]$ 

$\text{=}\dfrac{\text{1}}{\left( \text{2k+3} \right)}\left[ \dfrac{\text{2}{{\text{k}}^{\text{2}}}\text{+5k+3}}{\text{3}\left( \text{2k+5} \right)} \right]$ 

$\text{=}\dfrac{\text{1}}{\left( \text{2k+3} \right)}\left[ \dfrac{\text{2}{{\text{k}}^{\text{2}}}\text{+2k+3k+3}}{\text{3}\left( \text{2k+5} \right)} \right]$ 

$\text{=}\dfrac{\text{1}}{\left( \text{2k+3} \right)}\left[ \dfrac{\text{2k}\left( \text{k+1} \right)\text{+3}\left( \text{k+1} \right)}{\text{3}\left( \text{2k+5} \right)} \right]$ 

$\text{=}\dfrac{\left( \text{k+1} \right)\left( \text{2k+3} \right)}{\text{3}\left( \text{2k+3} \right)\left( \text{2k+5} \right)}$ 

$\text{=}\dfrac{\left( \text{k+1} \right)}{\text{3}\left\{ \text{2}\left( \text{k+1} \right)\text{+3} \right\}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

18. Prove the following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

\[\text{1+2+3+}...\text{+n}\dfrac{\text{1}}{\text{8}}{{\left( \text{2n+1} \right)}^{\text{2}}}\] 

Ans: Let us denote the given equality by $\text{P(n),}$ i.e.,

$\text{P}\left( \text{n} \right)\text{:1+2+3+}...\text{+n}\dfrac{\text{1}}{\text{8}}{{\left( \text{2n+1} \right)}^{\text{2}}}$ 

For $\text{n=1}$,

$\text{1}\dfrac{\text{1}}{\text{8}}{{\left( \text{2}\text{.1+1} \right)}^{\text{2}}}\text{=}\dfrac{\text{9}}{\text{8}}$

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some positive integer $\text{k}$, i.e.,

$\text{1+2+3+}...\text{+k}\dfrac{\text{1}}{\text{8}}{{\left( \text{2k+1} \right)}^{\text{2}}}...\text{(i)}$ 

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true.

Consider

$\text{1+2+3+}...\text{+k+}\left( \text{k+1} \right)\text{}\dfrac{\text{1}}{\text{8}}{{\left( \text{2k+1} \right)}^{\text{2}}}\text{+}\left( \text{k+1} \right)$        [Using (i)]

$\text{}\dfrac{\text{1}}{\text{8}}\left\{ {{\left( \text{2k+1} \right)}^{\text{2}}}\text{+8}\left( \text{k+1} \right) \right\}$ 

$\text{}\dfrac{\text{1}}{\text{8}}\left\{ \text{4}{{\text{k}}^{\text{2}}}\text{+4k+1+8k+8} \right\}$ 

$\text{}\dfrac{\text{1}}{\text{8}}\left\{ \text{4}{{\text{k}}^{\text{2}}}\text{+12k+9} \right\}$ 

$\text{}\dfrac{\text{1}}{\text{8}}{{\left( \text{2k+3} \right)}^{\text{2}}}$ 

$\text{}\dfrac{\text{1}}{\text{8}}{{\left\{ \text{2}\left( \text{k+1} \right)\text{+1} \right\}}^{\text{2}}}$ 

Hence, $\text{1+2+3+}...\text{+k+}\left( \text{k+1} \right)\text{}\dfrac{\text{1}}{\text{8}}{{\left( \text{2k+1} \right)}^{\text{2}}}\text{+}\left( \text{k+1} \right)$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

19. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{n}\left( \text{n+1} \right)\left( \text{n+5} \right)$is a multiple of $\text{3}$.

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right):\text{n}\left( \text{n+1} \right)\left( \text{n+5} \right)$, which is a multiple of $3$.

For $\text{n=1}$,

$\text{1}\left( \text{1+1} \right)\left( \text{1+5} \right)\text{=12}$, which is a multiple of $3$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

$\text{k}\left( \text{k+1} \right)\left( \text{k+5} \right)$ is a multiple of $3$.

$\therefore \text{k}\left( \text{k+1} \right)\left( \text{k+5} \right)\text{=3m}$, where $\text{m}\in \text{N}$ …(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

$\left( \text{k+1} \right)\left\{ \left( \text{k+1} \right)\text{+1} \right\}\left\{ \left( \text{k+1} \right)\text{+5} \right\}$ 

$\text{=}\left( \text{k+1} \right)\left( \text{k+2} \right)\left\{ \left( \text{k+1} \right)\text{+5} \right\}$ 

$\text{=}\left( \text{k+1} \right)\left( \text{k+2} \right)\left( \text{k+5} \right)\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)$ 

$\text{=}\left\{ \text{k}\left( \text{k+1} \right)\left( \text{k+5} \right)\text{+2}\left( \text{k+1} \right)\left( \text{k+5} \right) \right\}\text{+}\left( \text{k+1} \right)\left( \text{k+2} \right)$ 

$\text{=3m+}\left( \text{k+1} \right)\left\{ \text{2}\left( \text{k+5} \right)\text{+}\left( \text{k+2} \right) \right\}$ 

$\text{=3m+}\left( \text{k+1} \right)\left\{ \text{2k+10+k+2} \right\}$ 

$\text{=3m+}\left( \text{k+1} \right)\left\{ \text{3k+12} \right\}$ 

$\text{=3m+3}\left( \text{k+1} \right)\left\{ \text{k+4} \right\}$ 

$\text{=3}\left\{ \text{m+}\left( \text{k+1} \right)\left( \text{k+4} \right) \right\}\text{=3 }\!\!\times\!\!\text{ q}$, where $\text{q=}\left\{ \text{m+}\left( \text{k+1} \right)\left( \text{k+4} \right) \right\}$ is some natural number.

Hence, $\left( \text{k+1} \right)\left\{ \left( \text{k+1} \right)\text{+1} \right\}\left\{ \left( \text{k+1} \right)\text{+5} \right\}$ is a multiple of $\text{3}$.

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

20. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{1}{{\text{0}}^{\text{2n-1}}}\text{+1}$ is divisible by $\text{11}$ 

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right)\text{:1}{{\text{0}}^{\text{2n-1}}}\text{+1}$ is divisible by $\text{11}$.

For $\text{n=1}$,

$\text{P}\left( 1 \right)\text{=1}{{\text{0}}^{\text{2n-1}}}\text{+1=11}$ and $\text{P}\left( 1 \right)$ is divisible by $\text{11}$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

i.e., $\text{1}{{\text{0}}^{\text{2k-1}}}\text{+1}$ is divisible by $\text{11}$.

$\therefore \text{1}{{\text{0}}^{\text{2k-1}}}\text{+1=11m}$, where $\text{m}\in \text{N}$…(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

$\text{1}{{\text{0}}^{\text{2}\left( \text{k+1} \right)\text{-1}}}\text{+1}$ 

$\text{=1}{{\text{0}}^{\text{2k+2-1}}}\text{+1}$ 

$\text{=1}{{\text{0}}^{\text{2k+1}}}\text{+1}$ 

$\text{=1}{{\text{0}}^{\text{2}}}\left( \text{1}{{\text{0}}^{\text{2k-1}}}\text{+1-1} \right)\text{+1}$ 

$\text{=1}{{\text{0}}^{\text{2}}}\left( \text{1}{{\text{0}}^{\text{2k-1}}}\text{+1} \right)\text{-1}{{\text{0}}^{\text{2}}}\text{+1}$

$\text{=1}{{\text{0}}^{\text{2}}}\text{.11m-100+1}$              [Using (i)]

$\text{=100 }\!\!\times\!\!\text{ 11m-99}$ 

$\text{=11}\left( \text{100m-9} \right)$ 

$\text{=11r}$, where $\text{r=}\left( \text{100m-9} \right)$ is some natural number

Therefore, $\text{1}{{\text{0}}^{\text{2}\left( \text{k+1} \right)\text{-1}}}\text{+1}$ is divisible by $\text{11}$.

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

21. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

${{\text{x}}^{\text{2n}}}\text{-}{{\text{y}}^{\text{2n}}}$ is divisible by $\text{x+y}$.

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right)\text{:}{{\text{x}}^{\text{2n}}}\text{-}{{\text{y}}^{\text{2n}}}$ is divisible by $\text{x+y}$.

For $\text{n=1}$,

$\text{P}\left( 1 \right)\text{=}{{\text{x}}^{\text{2 }\!\!\times\!\!\text{ 1}}}\text{-}{{\text{y}}^{\text{2 }\!\!\times\!\!\text{ 1}}}\text{=}{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=}\left( \text{x+y} \right)\left( \text{x-y} \right)$, which is clearly divisible by $\text{x+y}$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

${{\text{x}}^{\text{2k}}}\text{-}{{\text{y}}^{\text{2k}}}$ is divisible by $\text{x+y}$.

$\therefore $ Let ${{\text{x}}^{\text{2k}}}\text{-}{{\text{y}}^{\text{2k}}}\text{=m}\left( \text{x+y} \right)$, where $\text{m}\in \text{N}$…(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

\[{{\text{x}}^{\text{2}\left( \text{k+1} \right)\text{-}{{\text{y}}^{\text{2}\left( \text{k+1} \right)}}}}\] 

$\text{=}{{\text{x}}^{\text{2k}}}\text{.}{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2k}}}\text{.}{{\text{y}}^{\text{2}}}$ 

$\text{=}{{\text{x}}^{\text{2}}}\left( {{\text{x}}^{\text{2k}}}\text{-}{{\text{y}}^{\text{2k}}}\text{+}{{\text{y}}^{\text{2k}}} \right)\text{-}{{\text{y}}^{\text{2k}}}\text{.}{{\text{y}}^{\text{2}}}$ 

\[\text{=}{{\text{x}}^{\text{2}}}\left\{ \text{m}\left( \text{x+y} \right)\text{+}{{\text{y}}^{\text{2k}}} \right\}\text{-}{{\text{y}}^{\text{2k}}}\text{.}{{\text{y}}^{\text{2}}}\] 

\[\text{=m}\left( \text{x+y} \right){{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2k}}}\text{.}{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2k}}}\text{.}{{\text{y}}^{\text{2}}}\] 

\[\text{=m}\left( \text{x+y} \right){{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2k}}}\left( {{\text{x}}^{2}}\text{-}{{\text{y}}^{2}} \right)\] 

\[\text{=m}\left( \text{x+y} \right){{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2k}}}\left( \text{x+y} \right)\left( \text{x-y} \right)\] 

$\text{=}\left( \text{x+y} \right)\left\{ \text{m}{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2k}}}\left( \text{x-y} \right) \right\}$, which is a factor of $\left( \text{x+y} \right)$.

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

22. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

${{\text{3}}^{\text{2n+2}}}\text{-8n-9}$ is divisible by $\text{8}$. 

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right)\text{:}{{\text{3}}^{\text{2n+2}}}\text{-8n-9}$ is divisible by $\text{8}$.

For $\text{n=1}$,

$\text{P}\left( \text{n} \right)={{\text{3}}^{\text{2 }\!\!\times\!\!\text{ 1+2}}}\text{-8 }\!\!\times\!\!\text{ 1-9=64}$, which is divisible by $\text{8}$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

${{\text{3}}^{\text{2k+2}}}\text{-8k-9}$ is divisible by $\text{8}$.

$\therefore {{\text{3}}^{\text{2k+2}}}\text{-8k-9=8m}$; where $\text{m}\in \text{N}$…(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

${{\text{3}}^{\text{2}\left( \text{k+1} \right)\text{+2}}}\text{-8}\left( \text{k+1} \right)\text{-9}$ 

${{\text{3}}^{\text{2k+2}}}\text{.}{{\text{3}}^{\text{2}}}\text{-8k-8-9}$ 

$\text{=}{{\text{3}}^{\text{2}}}\left( {{\text{3}}^{\text{2k+2}}}\text{-8k-9+8k+9} \right)\text{-8k-17}$ 

$\text{=}{{\text{3}}^{\text{2}}}\left( {{\text{3}}^{\text{2k+2}}}\text{-8k-9} \right)\text{+}{{\text{3}}^{\text{2}}}\left( \text{8k+9} \right)\text{-8k-17}$ 

$\text{=9}\text{.8m+9}\left( \text{8k+9} \right)\text{-8k-17}$ 

$\text{=9}\text{.8m+72k+81-8k-17}$ 

$\text{=9}\text{.8m+64k+64}$ 

$\text{=8}\left( \text{9m+8k+8} \right)$ 

$\text{=8r}$, where $\text{r=}\left( \text{9m+8k+8} \right)$ is a natural number

Therefore, ${{\text{3}}^{\text{2}\left( \text{k+1} \right)\text{+2}}}\text{-8}\left( \text{k+1} \right)\text{-9}$ is divisible by $\text{8}$.

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

23. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\text{4}{{\text{1}}^{\text{n}}}\text{-1}{{\text{4}}^{\text{n}}}$ is a multiple of $\text{27}$.

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right)\text{:4}{{\text{1}}^{\text{n}}}\text{-1}{{\text{4}}^{\text{n}}}$ is a multiple of $27$.

It can be observed that $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$

For $\text{n=1}$,

$\text{P}\left( 1 \right)={{41}^{1}}-{{14}^{1}}=27$, which is a multiple of $27$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

$\text{4}{{\text{1}}^{\text{k}}}\text{-1}{{\text{4}}^{\text{k}}}$ is a multiple of $27$.

$\therefore \text{4}{{\text{1}}^{\text{k}}}\text{-1}{{\text{4}}^{\text{k}}}\text{=27m,}\,\,\,\text{m}\in \text{N}$ …(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

$\text{4}{{\text{1}}^{\text{k+1}}}\text{-1}{{\text{4}}^{\text{k+1}}}$ 

$\text{=4}{{\text{1}}^{\text{k}}}\text{.41-1}{{\text{4}}^{\text{k}}}\text{.14}$ 

$\text{=41}\left( \text{4}{{\text{1}}^{\text{k}}}\text{-1}{{\text{4}}^{\text{k}}}\text{+1}{{\text{4}}^{\text{k}}} \right)\text{-1}{{\text{4}}^{\text{k}}}\text{.14}$ 

$\text{=41}\text{.27m+1}{{\text{4}}^{\text{k}}}\left( \text{41-14} \right)$ 

$\text{=41}\text{.27m+27}\text{.1}{{\text{4}}^{\text{k}}}$ 

$\text{=27}\left( \text{41m-1}{{\text{4}}^{\text{k}}} \right)$ 

$\text{=27 }\!\!\times\!\!\text{ r}$, where $\text{r=}\left( \text{41m-1}{{\text{4}}^{\text{k}}} \right)$ is a natural number.

Therefore, $\text{4}{{\text{1}}^{\text{k+1}}}\text{-1}{{\text{4}}^{\text{k+1}}}$ is a multiple of $27$.

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

24. Prove that following by using the principle of mathematical induction for all $\text{n}\in \text{N}$:

$\left( \text{2n+7} \right)\text{}{{\left( \text{n+3} \right)}^{\text{2}}}$ 

Ans: Let us denote the given statement by $\text{P}\left( \text{n} \right)$, i.e.,

$\text{P}\left( \text{n} \right)\text{:}\left( \text{2n+7} \right)\text{}{{\left( \text{n+3} \right)}^{\text{2}}}$ 

For $\text{n=1}$,

$\text{P}\left( 1 \right)\text{=2}\text{.1+7=9}{{\left( \text{1+3} \right)}^{\text{2}}}\text{=16}$, which is true because $\text{916}$.

Therefore, $\text{P}\left( \text{n} \right)$ is true for $\text{n=1}$.

Let us assume that $\text{P}\left( \text{k} \right)$ is true for some natural number $\text{k}$, i.e.,

$\left( \text{2k+7} \right)\text{}{{\left( \text{k+3} \right)}^{\text{2}}}$ …(i)

Now, we have to prove that $\text{P}\left( \text{k+1} \right)$ is also true whenever $\text{P}\left( \text{k} \right)$ is true.

Consider

$\left\{ \text{2}\left( \text{k+1} \right)\text{+7} \right\}\text{=}\left( \text{2k+7} \right)\text{+2}$ 

$\therefore \left\{ \text{2}\left( \text{k+1} \right)\text{+7} \right\}\text{=}\left( \text{2k+7} \right)\text{+2}{{\left( \text{k+3} \right)}^{\text{2}}}\text{+2}$      [Using (i)]

$\text{2}\left( \text{k+1} \right)\text{+7}{{\text{k}}^{\text{2}}}\text{+6k+9+2}$ 

$\text{2}\left( \text{k+1} \right)\text{+7}{{\text{k}}^{\text{2}}}\text{+6k+11}$ 

Now, ${{\text{k}}^{\text{2}}}\text{+6k+11}{{\text{k}}^{\text{2}}}\text{+8k+16}$ 

$\therefore \text{2}\left( \text{k+1} \right)\text{+7}{{\left( \text{k+4} \right)}^{\text{2}}}$ 

$\text{2}\left( \text{k+1} \right)\text{+7}{{\left\{ \left( \text{k+1} \right)\text{+3} \right\}}^{\text{2}}}$ 

Therefore, $\text{P}\left( \text{k+1} \right)$ holds whenever $\text{P}\left( \text{k} \right)$ holds.

Hence, the given equality is true for all natural numbers i.e., $\text{N}$ by the principle of mathematical induction.

Chapter 4 Principle of Mathematical Induction NCERT Solutions

CBSE Class 11 Mathematics spans over a very vast syllabus and covers some very crucial topics that you may encounter in your entrance exams. Among them, the Principle of Mathematical Induction Class 11 can come across as a fairly difficult chapter for many students. However, it can be a scoring topic if the fundamental concepts are understood in detail by students.

Principle of Mathematical Induction Class 11 NCERT Solutions has been framed by teachers who understand the importance of step-by-step explanations for the maths questions based on the application of Mathematical Induction. Students can opt for NCERT Solutions for Class 11 Maths Chapter 4 PDF Download for seamless offline studying.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 4 Exercise

NCERT Solutions for Class 11 Maths Chapter 4 Sub-topics

A brief summary of the topics and sub-topics covered in the chapter of Mathematical Induction Class 11 is given below.

Part 1: Introduction

The first part of the chapter imparts a basic idea of deductive reasoning. Questions from this section consist of two or more sentences stating an argument or a logical question that students will need to prove as true or false with the help of derived deductive steps.

In order to gain a clearer understanding of the theorems involved in deriving deductive statements, you can refer to NCERT Solutions for Class 11 Maths Chapter 4 PDF. You will find a detailed breakdown of the application of principles that will help you to solve the sums given in this chapter.

Part 2: Motivation

In this section, students will learn about an important concept that they need to understand before solving the sums of Mathematical Induction. In Mathematical Induction Class 11 Solutions, the principle of Motivation involves the process of proving that if a given statement is true for one natural number, then it also holds true for the rest of n natural numbers. This lays the base for the first principle of Mathematical Induction. Hence it is very important that you develop a strong grip on this topic in order to solve related sums.

Part 3: Illustration

The next section teaches students about the process of deriving an equation. However, this is an area where many students struggle to make use of the right logic to prove the given statements. At this time, the Principle of Mathematical Induction NCERT Solutions comes in handy to help students find the correct method of proceeding with deducting statements for equations.

Part 4: The Principle of Mathematical Induction

To round up, the final drill in the chapter these NCERT Solutions will guide you through understanding the principle of Mathematical Induction with the help of the previously discussed concepts. You will, thus, be required to solve equations based on applications of the same.

For better guidance and thorough understanding, the Principle of Mathematical Induction Class 11 Solutions PDF comes with detailed explanations to help students fathom the questions correctly, and then proceeds with the step-wise solutions, which have again been broken down with explanatory notes.

How can you benefit from the Principle of Mathematical Induction Class 11 Solutions?

Chapter 4 Maths Class 11 might come off as a remarkably complex chapter but not for those who address and clear their troubles with the basics first. For that, a thorough acquaintance with the prescribed textbooks is undoubtedly very important. What can help additionally are PMI Class 11 NCERT Solutions as it can guide you through key concepts with smart problem-solving techniques and elaborate answers to help you self-study and practice the sums.