NCERT Solutions for Class 11 Maths Chapter 4 - Exercise

EXERCISE 4.1

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

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

Ans: Let the given statement be $P(n)$, i.e., $P(n):1 + 3 + {3^2} +  \ldots .. + {3^{ n  - 1}} = \dfrac{{\left( {{3^n} - 1} \right)}}{2}$

For $n = 1$ we have $P(1): = \dfrac{{\left( {{3^1} - 1} \right)}}{2} = \dfrac{{3 - 1}}{2} = \dfrac{2}{2} = 1$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $1 + 3 + {3^2} +  \ldots  + {3^{k - 1}} = \dfrac{{\left( {{3^k} - 1} \right)}}{2}$

We shall now prove that $P(k + 1)$ is true. Consider $1 + 3 + {3^2} +  \ldots  + {3^{k - 1}} + {3^{(k + 1) - 1}}$

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

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

$ = \dfrac{{\left( {{3^k} - 1} \right) + {{2.3}^k}}}{2}$

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

$ = \dfrac{{{{3.3}^k} - 1}}{2}$

$ = \dfrac{{{3^{k + 1}} - 1}}{2}$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., ${\text{N}}$.

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

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

Ans: Let the given statement be $P(n)$, i.e.. $P(n):{1^3} + {2^3} + {3^3} +  \ldots .. + {n^3} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}$

For $n = 1$, we have $P(1):{1^3} = 1 = \left( {\dfrac{{1{{(1 + 1)}^2}}}{2}} \right) = {\left( {\dfrac{{1.2}}{2}} \right)^2} = {1^2} = 1$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., ${1^3} + {2^2} + {3^3} +  \ldots  +  + {k^3} - {\left( {\dfrac{{k(k + 1)}}{2}} \right)^2} \ldots  \ldots $

We shall now prove that $P(k + 1)$ is true. Consider ${1^3} + {2^3} + {3^3} +  \ldots  \ldots  + {k^3} + {(k + 1)^3}$

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

$ = {\left( {\dfrac{{k(k + 1)}}{2}} \right)^2} + {(k + 1)^2}\quad [U\operatorname{sing} (i)]$

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$1 + \dfrac{1}{{(1 + 2)}} + \dfrac{1}{{(1 + 2 + 3)}} +  \ldots . + \dfrac{1}{{(1 + 2 + 3 +  \ldots n)}} = \dfrac{{2n}}{{(n + 1)}}$

Ans: Let the given statement be $P(n)$. i.e.$P(n):1 + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} +  \ldots .. + \dfrac{1}{{1 + 2 + 3 +  \ldots n}} - \dfrac{{2n}}{{n + 1}}$

For $n = 1$, we have $P(1):1 = \dfrac{{2.1}}{{1 + 1}} = \dfrac{2}{2} = 1$ which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $1 + \dfrac{1}{{1 + 2}} +  \ldots  + \dfrac{1}{{1 + 2 + 3}} +  \ldots  + \dfrac{1}{{1 + 2 + 3 +  \ldots . + k}} = \dfrac{{2k}}{{k + 1}}.....(i)$

We shall now prove that $P(k + 1)$ is true. 

Consider, 

$1 + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} +  \ldots .. + \dfrac{1}{{1 + 2 + 3 +  \ldots .. + k}} + \dfrac{1}{{1 + 2 + 3 +  \ldots .. + k + (k + 1)}}$

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

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$1.2.3 + 2.3.4 +  \ldots  + n(n + 1)(n + 2) = \dfrac{{n(n + 1)(n + 2)(n + 3)}}{4}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):1.2.3 + 2.3.4 +  \ldots  + n(n + 1)(n + 2) = \dfrac{{n(n + 1)(n + 2)(n + 3)}}{4}$

For $n = 1$, we have $P(1):1.2.3 = 6 = \dfrac{{1(1 + 1)(1 + 2)(1 + 3)}}{4} = \dfrac{{1.2 \cdot 3.4}}{4} = 6$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $1.2.3 + 2.3 \cdot 4 +  \ldots .. + k(k + 1)(k + 2) = \dfrac{{k(k + 1)(k + 2)(k + 3)}}{4}$

We shall now prove that $P(k + 1)$ is true. Consider $1.2.3 + 2.3.4 +  \ldots .. + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)$

$ = \{ 1.2.3 + 2.3.4 +  \ldots .. + k(k + 1)(k + 2)\}  + (k + 1)(k + 2) + (k + 3)$

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

5. Prove the following by using the principle of mathematical induction for all $n \in N$ :

$1.3 + {2.3^2} + {3.3^3} +  \ldots . + n \cdot {3^n} = \dfrac{{(2n - 1){3^{n + 1}} + 3}}{4}$

Ans: Let the given statement be $P(n)$, i.e.,

$P(n):1.3 + {2.3^2} + {3.3^7} +  \ldots  + n{3^n} = \dfrac{{(2n - 1){3^{n + 1}} + 3}}{4}$

For $n = 1$, we have $P(1):1.3 = 3 = \dfrac{{(2.1 - 1){3^{1 - 1}} + 3}}{4} = \dfrac{{{3^2} + 3}}{4} = \dfrac{{12}}{4} = 3$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $1.3 + {23^2} + {33^3} +  \ldots . + k{3^k} = \dfrac{{(2k - 1){3^{k + 1}} + 3}}{4}$

We shall now prove that $P(k + 1)$ is true. Consider $1.3 + {2.3^2} + {3.3^3} +  \ldots . + k \cdot {3^k} + (k + 1) \cdot {3^{k + 1}}$

$ = \left( {1.3 + 2 \cdot {3^2} + {{3.3}^3} +  \ldots .. + k \cdot {3^k}} \right) + (k + 1) \cdot {3^{x + 1}}$

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

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

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

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

$ = \dfrac{{{3^{k + 1}} \cdot 3\{ 2k + 1\}  + 3}}{4}$

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement ${\text{P}}({\text{n}})$ is true for all natural numbers i.e., ${\text{N}}$.

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

$1.2 + 2.3 + 3.4 +  \ldots  + n(n + 1) = \left[ {\dfrac{{n(n + 1)(n + 2)}}{3}} \right]$

Ans: Let the given statement be $P(n)$, i.e.,

$P(n):1.2 + 2.3 + 3.4 +  \ldots  + n.(n + 1) = \left[ {\dfrac{{n(n + 1)(n + 2)}}{3}} \right]$

For $n = 1$, we have $P(1):1.2 = 2 = \dfrac{{1(1 + 1)(1 + 2)}}{3} = \dfrac{{1.2.3}}{3} = 2$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $1.2 + 2.3 + 3.4 +  \ldots .. + k \cdot (k + 1) = \left[ {\dfrac{{k(k + 1)(k + 2)}}{3}} \right] \ldots  \ldots ..(i)$

We shall now prove that $P(k + 1)$ is true. Consider $1.2 + 2 \cdot 3 + 3 \cdot 4 +  \ldots  + {k_ - }(k + 1) + (k + 1) \cdot (k + 2)$

$ = [12 + 2.3 + 3.4 +  \ldots .. + k \cdot (k + 1)] + (k + 1).(k + 2)$

$ = \dfrac{{k(k + 1)(k + 2)}}{3} + (k + 1)(k + 2)\quad [U\operatorname{sing} (i)]$

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$1.3 + 3.5 + 5.7 +  \ldots .. + (2n - 1)(2n + 1) = \dfrac{{n\left( {4{n^2} + 6n - 1} \right)}}{3}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):1 \cdot 3 + 3.5 + 5.7 +  \ldots  + (2n - 1)(2n + 1) = \dfrac{{n\left( {4{n^2} + 6n - 1} \right)}}{3}$

For $n = 1$, we have $P(1):1.3 = 3 = \dfrac{{1\left( {{{4.1}^2} + 6.1 - 1} \right)}}{3} = \dfrac{{4 + 6 - 1}}{3} = \dfrac{9}{3} = 3$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.,

$1.3 + 3.5 + 5.7 +  \ldots . + (2k - 1)(2k + 1) = \dfrac{{k\left( {4{k^2} + 6k - 1} \right)}}{3}.....(i)$

We shall now prove that $P(k + 1)$ is true. Consider $(1.3 + 3.5 + 5.7 +  \ldots .. + (2k - 1)(2k + 1)) + \{ (k + 1) - 1\} \{ 2(k + 1) + 1\} $

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

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

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

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

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

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$1.2 + 2 \cdot {2^2} + {3.2^2} +  \ldots  + n \cdot {2^n} = (n - 1){2^{n + 1}} + 2$

Ans: Let the given statement be $P(n)$, i.e., $P(n):1.2 + 2 \cdot {2^2} + {32^2} +  \ldots  + n{2^2} = (n - 1){2^{n + 1}} + 2$

For $n = 1$, we have $P(1):1.2 = 2 = (1 - 1){2^{1 + 1}} + 2 = 0 + 2 = 2$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.. $1.2 + {2.2^2} + {3.2^2} +  \ldots  + k \cdot {2^k} = (k - 1){2^{k + 1}} + 2 \ldots .(i)$

We shall now prove that $P(k + 1)$ is true. Consider $\left\{ {1.2 + {{2.2}^2} + {{3.2}^2} +  \ldots . + k \cdot {2^*}} \right\} + (k + 1) \cdot {2^{k + 1}}$

$ = (k - 1){2^{k + t}} + 2 + (k + 1){2^{k - 2}}$

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

$ = {2^{k + 1}} \cdot 2k + 2$

$ = k \cdot {2^{(k + 1) + 1}} + 2$

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  \ldots  \ldots  + \dfrac{1}{{{2^n}}} - 1 - \dfrac{1}{{{2^n}}}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  \ldots .. + \dfrac{1}{{{2^n}}} = 1 - \dfrac{1}{{{2^8}}}$

For $n = 1$, we have $P(1):\dfrac{1}{2} = 1 - \dfrac{1}{{{2^1}}} = \dfrac{1}{2}$, which is true.

Let $P(k)$ be true for some positive integer ${k_1}$ i.e., $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  \ldots . + \dfrac{1}{{{2^k}}} = 1 - \dfrac{1}{{{2^k}}} \ldots ..(i)$

We shall now prove that $P(k + 1)$ is true. Consider $\left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  \ldots .. + \dfrac{1}{{{2^k}}}} \right) + \dfrac{1}{{{2^{k - 2}}}}$

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

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

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

$ = 1 - \dfrac{1}{{{2^{k - t}}}}$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$\dfrac{1}{{2.5}} + \dfrac{1}{{5.8}} + \dfrac{1}{{8.11}} +  \ldots .. + \dfrac{1}{{(3n - 1)(3n + 2)}} = \dfrac{n}{{(6n + 4)}}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):\dfrac{1}{{2.5}} + \dfrac{1}{{5.8}} + \dfrac{1}{{8.11}} +  \ldots  \ldots  + \dfrac{1}{{(3n - 1)(3n + 2)}} = \dfrac{n}{{(6n + 4)}}$

For $n = 1$, we have $P(1) = \dfrac{1}{{2.5}} = \dfrac{1}{{10}} = \dfrac{1}{{6.1 + 4}} = \dfrac{1}{{10}}$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.. $\dfrac{1}{{2.5}} + \dfrac{1}{{5.8}} + \dfrac{1}{{8.11}} +  \ldots .. + \dfrac{1}{{(3k + 1)(3k + 2)}} = \dfrac{k}{{6k + 4}} \ldots ..(i)$

We shall now prove that $P(k + 1)$ is true. Consider $\dfrac{1}{{2.5}} + \dfrac{1}{{5.8}} + \dfrac{1}{{8.11}} +  \ldots . + \dfrac{1}{{(3k + 1)(3k + 2)}} + \dfrac{1}{{\{ 3(k + 1) - 1\} \{ 3(k + 1) + 2\} }}$

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

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

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e, ${\text{N}}$.

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

$\dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3 \cdot 4}} + \dfrac{1}{{3.4.5}} +  \ldots . + \dfrac{1}{{n(n + 1)(n + 2)}} = \dfrac{{n(n + 3)}}{{4(n + 1)(n + 2)}}$

Ans: Let the given statement be $P(n)$, i.e.. $P(n):\dfrac{1}{{12.3}} + \dfrac{1}{{23.4}} + \dfrac{1}{{3.4.5}} +  \ldots . + \dfrac{1}{{n(n + 1)(n + 2)}} = \dfrac{{n(n + 3)}}{{4(n + 1)(n + 2)}}$

For $n = 1$, we have $P(1) = \dfrac{1}{{1 \cdot 2 - 3}} - \dfrac{{1 \cdot (1 + 3)}}{{4(1 + 1)(1 + 2)}} = \dfrac{{1 \cdot 4}}{{4 \cdot 2 \cdot 3}} - \dfrac{1}{{1 \cdot 2 - 3}}$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.. $\dfrac{1}{{1.2 \cdot 3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} +  \ldots  \ldots  + \dfrac{1}{{k(k + 1)(k + 2)}} = \dfrac{{k(k + 3)}}{{4(k + 1)(k + 2)}}.....(i)$

We shall now prove that $P(k + 1)$ is true. Consider $\left[ {\dfrac{1}{{1 \cdot 2 - 3}} + \dfrac{1}{{2 \cdot 3 - 4}} + \dfrac{1}{{3 \cdot 4 \cdot 5}} +  \ldots .. + \dfrac{1}{{k(k + 1)(k + 2)}}} \right] + \dfrac{1}{{(k + 1)(k + 2)(k + 3)}}$

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

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

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

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

12. Prove the following by using the principle of mathematical induction for all 

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

Ans: Let the given statement be $P(n)$, i.e.. $P(n):a + ar + a{r^2} +  \ldots .. + a{r^{n - 1}} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$

For $n = 1$, we have $P(1):a = \dfrac{{a\left( {{r^1} - 1} \right)}}{{(r - 1)}} = a$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., ${\text{a}} + {\text{ar}} + {\text{a}}{{\text{r}}^2} +  \ldots  \ldots  + {\text{a}}{{\text{r}}^{k - 1}} = \dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{k}}} - 1} \right)}}{{{\text{r}} - 1}} \ldots  \ldots ({\text{i}})$

We shall now prove that $P(k + 1)$ is true.

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

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

Ans: Let the given statement be $P(n)$, i.e.. $P(n):\left( {1 + \dfrac{3}{1}} \right)\left( {1 + \dfrac{5}{4}} \right)\left( {1 + \dfrac{7}{9}} \right) \ldots \left( {1 + \dfrac{{(2n + 1)}}{{{n^2}}}} \right) = {(n + 1)^2}$

For $n = 1$, we have $P(1):\left( {1 + \dfrac{3}{1}} \right) = 4 = {(1 + 1)^2} = {2^2} = 4$, which is true.

Let $P(k)$ be true for some positive integer ${k_t}$ i.e.. $\left( {1 + \dfrac{3}{1}} \right)\left( {1 + \dfrac{5}{4}} \right)\left( {1 + \dfrac{7}{9}} \right) \ldots .\left( {1 + \dfrac{{(2k + 1)}}{{{k^2}}}} \right) = {(k + 1)^2} \ldots  \ldots (1)$

We shall now prove that $P(k + 1)$ is true. Consider $\left[ {\left( {1 + \dfrac{3}{1}} \right)\left( {1 + \dfrac{5}{4}} \right)\left( {1 + \dfrac{7}{9}} \right) \ldots \left( {1 + \dfrac{{(2k + 1)}}{{{k^2}}}} \right)} \right]\left\{ {1 + \dfrac{{\{ 2(k + 1) + 1\} }}{{{{(k + 1)}^2}}}} \right\}$

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

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

$ = {(k + 1)^2} + 2(k + 1) + 1$

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

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

Ans: Let the given statement be $P(n)$, i.e.. $P(n):\left( {1 + \dfrac{1}{1}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{3}} \right) \ldots .\left( {1 + \dfrac{1}{n}} \right) = (n + 1)$

For $n = 1$, we have $P(1):\left( {1 + \dfrac{1}{1}} \right) = 2 = (1 + 1)$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $P(k):\left( {1 + \dfrac{1}{1}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{3}} \right) \ldots \left( {1 + \dfrac{1}{k}} \right) = (k + 1) \ldots  \ldots .(1)$

We shall now prove that $P(k + 1)$ is true. Consider $\left[ {\left( {1 + \dfrac{1}{1}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{3}} \right) \ldots \left( {1 + \dfrac{1}{k}} \right)} \right]\left( {1 + \dfrac{1}{{k + 1}}} \right)$

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

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

$ = (k + 1) + 1$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

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

Ans: Let the given statement be $P(n)$, i.e.. $P(n):{1^2} + {3^2} + {5^2} +  \ldots . + {(2n - 1)^2} = \dfrac{{n(2n - 1)(2n + 1)}}{3}$

For $n = 1$, we have $P(1) = {1^2} = 1 = \dfrac{{1(2.1 - 1)(21 + 1)}}{3} = \dfrac{{1.1.3}}{3} = 1$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.. $P(k) = {1^2} + {3^2} + {5^2} +  \ldots  + {(2k - 1)^2} = \dfrac{{k(2k - 1)(2k + 1)}}{3}$

We shall now prove that $P(k + 1)$ is true. 

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

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

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

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

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$\dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} +  \ldots  + \dfrac{1}{{(3n - 2)(3n + 1)}} = \dfrac{n}{{(3n + 1)}}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):\dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7 \cdot 10}} +  \ldots  + \dfrac{1}{{(3n - 2)(3n + 1)}} = \dfrac{n}{{(3n + 1)}}$

For $n = 1$, we have $P(1) = \dfrac{1}{{1.4}} = \dfrac{1}{{3.1 + 1}} = \dfrac{1}{4} + \dfrac{1}{{1.4}}$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $P(k) = \dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} +  \ldots  + \dfrac{1}{{(3k - 2)(3k + 1)}} = \dfrac{k}{{3k + 1}}$

We shall now prove that $P(k + 1)$ is true. Consider $\left\{ {\dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} +  \ldots . + \dfrac{1}{{(3k - 2)(3k + 1)}}} \right\} + \dfrac{1}{{\{ 3(k + 1) - 2\} (3(k + 1) + 1\} }}$

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

${\text{[Using(1)]}}$

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

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

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

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

$ - \dfrac{{(3k + 1)(k + 1)}}{{(3k + 1)(3k + 4)}}$

$ - \dfrac{{(k + 1)}}{{3(k + 1) + 1}}$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$\dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + \dfrac{1}{{7.9}} +  \ldots  \ldots  + \dfrac{1}{{(2n + 1)(2n + 3)}} = \dfrac{n}{{3(2n + 3)}}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):\dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + \dfrac{1}{{7.9}} +  \ldots  \ldots  + \dfrac{1}{{(2n + 1)(2n + 3)}} = \dfrac{n}{{3(2n + 3)}}$

For $n = 1$, we have $P(1):\dfrac{1}{{3.5}} = \dfrac{1}{{3(2.1 + 3)}} = \dfrac{1}{{3.5}}$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e., $P(k):\dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + \dfrac{1}{{7.9}} +  \ldots . + \dfrac{1}{{(2k + 1)(2k + 3)}} = \dfrac{k}{{3(2k + 3)}} \ldots .(1)$

We shall now prove that $P(k + 1)$ is true. 

Consider $\left[ {\dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + \dfrac{1}{{7.9}} +  \ldots . + \dfrac{1}{{(2k + 1)(2k + 3)}}} \right] + \dfrac{1}{{\{ 2(k + 1) + 1\} \{ 2(k + 1) + 3\} }} = \dfrac{k}{{3(2k + 3)}} + \dfrac{1}{{(2k + 3)(2k + 5)}}$

$[{\text{Using(1)}}]$

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

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

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

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

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

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

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

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

18. Prove the following by using the principle of mathematical induction for all $n \in \mathbb{N}:$ $1 + 2 + 3 +  \ldots  + n < \dfrac{1}{8}{(2n + 1)^2}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):1 + 2 + 3 +  \ldots  + n < \dfrac{1}{8}{(2n + 1)^2}$

It can be noted that $P(n)$ is true for $n = 1$ since $1 < \dfrac{1}{8}{(2.1 + 1)^2} = \dfrac{9}{8}$

Let $P(k)$ be true for some positive integer $k$, i.e., $1 + 2 +  \ldots . + k < \dfrac{1}{8}{(2k + 1)^2} \ldots ..(1)$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider $(1 + 2 +  \ldots  + k) + (k + 1) < \dfrac{1}{8}{(2k + 1)^2} + (k + 1)\quad $ [Using (1)]

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

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

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

$ < \dfrac{1}{8}{(2k + 3)^2}$

$ < \dfrac{1}{8}{\{ 2(k + 1) + 1\} ^2}$

Hence, $(1 + 2 + 3 +  \ldots . + k) + (k + 1) < \dfrac{1}{8}{(2k + 1)^2} + (k + 1)$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$n(n + 1)(n + 5)$ is a multiple of 3

Ans: Let the given statement be $P(n)$, i.e.. $P(n):n(n + 1)(n + 5)$, which is a multiple of 3 .

It can be noted that $P(n)$ is true for $n = 1$ since $1(1 + 1)(1 + 5) = 12$, which is a multiple of 3.

Let $P(k)$ be true for some positive integer $k$, i.e., $k(k + 1)(k + 5)$ is a multiple of 3 $\therefore k(k + 1)(k + 5) = 3m$, where $m \in N$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider $(k + 1)\{ (k + 1) + 1\} \{ (k + 1) + 5\} $

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

$ = (k + 1)(k + 2)(k + 5) + (k + 1)(k + 2)$

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

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

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

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

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

$ = 3\{ m + (k + 1)(k + 4)\}  = 3 \times q$. where $q = \{ m + (k + 1)(k + 4)\} $ is some natural number. Therefore, $(k + 1)\{ (k + 1) + 1\} \{ (k + 1) + 5\} $ is a multiple of 3 . Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

${10^{2n - 1}} + 1$ is divisible by 11

Ans: Let the given statement be $P(n)$, i.e., $P(n):{10^{2n - 1}} + 1$ is divisible by 11 It can be observed that $P(n)$ is true for $n = 1$ Since $P(1) = {10^{2.1 - 1}} + 1 = 11$, which is divisible by 11 . Let $P(k)$ be true for some positive integer $k$,

i.e., ${10^{2k - 1}} + 1$ is divisible by 11 . $\therefore {10^{2k - 1}} + 1 = 11\;{\text{m}}$, where $m \in N$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider ${10^{2(k + 1) - 1}} + 1$

$ = {10^{2k + 2 - 1}} + 1$

$ = {10^{2k - 1}} + 1$

$ = {10^2}\left( {{{10}^{2t - 1}} + 1 - 1} \right) + 1$

$ = {10^2}\left( {{{10}^{2k - 1}} + 1} \right) - {10^2} + 1$

$ = {10^2}.11\;{\text{m}} - 100 + 1\quad [$ Using $(1)]$

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

$ = 11(100m - 9)$

$ = 11{\mathbf{r}}$, where $r = (100m - 9)$ is some natural number Therefore, ${10^{2(k + 1) - 1}} + 1$ is divisible by 11 . Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

${x^{2x}} - {y^{2 = }}$ is divisible by $x + y.$

Ans: Let the given statement be $P(n)$, i.e., $P(n):{x^{2n}} - {y^{2n}}$ is divisible by $x + y$ It can be observed that $P(n)$ is true for $n = 1$.

This is so because ${x^{2 - 1}} - {y^{2 - 4}} = {x^2} - {y^2} = (x + y)(x - y)$ is divisible by $(x + y)$. Let $P(k)$ be true for some positive integer ${k_1}$ i.e.. ${x^{3x}} - {y^{2k}}$ is divisible by $x + y$.

$\therefore $ Let ${x^{2k}} - {y^{2x}} = m(x + y)$, where $m \in N \ldots .(1)$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider ${x^{2x + 1\mid  - {y^4} - }}$

$ = {x^{2x}} - {x^2} - {y^{2x}} - {y^2}$

$ = {x^2}\left( {{x^{2k}} - {y^{2k}} + {y^{7x}}} \right) - {y^{2k}} - {y^2}$

$ = {x^2}\left\{ {m(x + y) + {y^{2k}}} \right\} - {y^{2k}} \cdot {y^2}\quad {\text{[Using(1)}}]$

$ = m(x + y){x^2} + {y^{2k}} \cdot {x^2} - {y^{2x}} - {y^2}$

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

$ = m(x + y){x^2} + {y^{2x}}(x + y)(x - y)$

$ = (x + y)\left\{ {m{x^2} + {y^{2k}}(x - y)} \right\}$, which is a factor of $(x + y)$. Thus, $P(k + 1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e.. ${\text{N}}$.

22. Prove the following by using the principle of mathematical induction for all $n \in \mathbb{N}:{3^{2n + 2}} - 8n - 9$ is divisible by 8

Ans: Let the given statement be $P(n)$, i.e., $P(n):{3^{2n + 2}} - 8n - 9$ is divisible by 8 . It can be observed that $P(n)$ is true for $n = 1$ Since ${3^{2 \cdot 1 - 2}} - 8 \times 1 - 9 = 64$, which is divisible by 8 . Let $P(k)$ be true for some positive integer

$k$, i.e, ${3^{2k + 2}} - 8k - 9$ is divisible by 8 . $\therefore {3^{2t + 2}} - 8k - 9 = 8\;{\text{ms}}$ where $m \in N \ldots  \ldots .(1)$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider ${3^{\alpha (k + 1) + 2}} - 8(k + 1) - 9$

${3^{2k + 2}} - {3^2} - 8k - 8 - 9$

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

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

$ = 9.8m + 9(8k + 9) - 8k - 17$

$ = 9.8m + 72k + 81 - 8k - 17$

$ = 9.8m + 64k + 64$

$ = 8(9m + 8k + 8)$

$ = 8r$, where $r = (9m + 8k + 8)$ is a natural number Therefore, ${3^{2(k - 2) - 2}} - 8(k + 1) - 9$ is divisible by 8 . Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all numbers i.e. ${\text{N}}$.

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

${41^n} - {14^n}$ is a multiple of 27.

Ans: Let the given statement be $P(n)$. i.e., $P(n):{41^*} - {14^*}$ is a multiple of 27

It can be observed that $P(n)$ is true for $n = 1$ Since ${41^1} - {14^1} = 27$, which is a multiple of 27. Let $P(k)$ be true for some positive integer $k$, i.e.. ${41^{\text{k}}} - {14^{\text{k}}}$ is a multiple of 27

$\therefore {41^k} - {14^k} = 27\;{\text{m}},m \in \mathbb{N} \ldots  \ldots (1)$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider ${41^{k + 1}} - {14^{k + 1}}$

$ = {41^k} \cdot 41 - {14^k} \cdot 14$

$ = 41\left( {{{41}^*} - {{14}^k} + {{14}^k}} \right) - {14^k} \cdot 14$

$ = 41.27\;{\text{m}} + {14^k}(41 - 14)$

$ = 41.27\;{\text{m}} + {27.14^{\text{k}}}$

$ = 27\left( {41m - {{14}^k}} \right)$

$ = 27 \times {\mathbf{r}}$, where $r = \left( {41m - {{14}^k}} \right)$ is a natural number. Therefore, ${41^{k + 1}} - {14^{k + 1}}$ is a multiple of 27 . Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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

$(2n + 7) < {(n + 3)^2}$

Ans: Let the given statement be $P(n)$, i.e., $P(n):(2n + 7) < {(n + 3)^2}$

It can be observed that $P(n)$ is true for $n = 1$ Since $2.1 + 7 = 9 < {(1 + 3)^2} = 16$, which is true. Let $P(k)$ be true for some positive integer ${k_1}$ i.e., $(2k + 7) < {(k + 3)^2} \ldots  \ldots (1)$

We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true. Consider $\{ 2(k + 1) + 7\}  = (2k + 7) + 2$

$\therefore \{ 2(k + 1) + 7\}  = (2k + 7) + 2 < {(k + 3)^2} + 2\quad [$ Using $(1)]$

$2(k + 1) + 7 < {k^2} + 6k + 9 + 2$

$2(k + 1) + 7 < {k^2} + 6k + 11$

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

$\therefore 2(k + 1) + 7 < {(k + 4)^2}$

$2(k + 1) + 7 < {\{ (k + 1) + 3\} ^2}$

Thus, $P(k + 1)$ is true whenever $P(k)$ is true. Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., ${\text{N}}$.

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 Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1

Opting for the NCERT solutions for Ex 4.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Principle of Mathematical Induction textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 4 Exercise 4.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 4 Exercise 4.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 4 Exercise 4.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.