NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series - Exercise 9.4

Exercise 9.4

1. Find the sum to $\mathbf{n}$ terms of the series  $\mathbf{1\times 2+2\times 3+3\times 4+4\times 5+}...$

Ans:

The series provided to us is $1\times 2+2\times 3+3\times 4+4\times 5+...$.

Note that, the ${{n}^{th}}$ term of the given series is ${{a}_{n}}=n\left( n+1 \right)$.

Therefore, the sum to the series up to $n$ terms,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{k\left( k+1 \right)} \\ & =\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{k} \\ \end{align}$

$\begin{align} & =\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{n\left( n+1 \right)}{2} \\ & =\frac{n\left( n+1 \right)}{2}\left( \frac{2n+1}{3}+1 \right) \\ & =\frac{n\left( n+1 \right)}{2}\left( \frac{2n+4}{3} \right) \\ \end{align}$

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

2. Find the sum to $\mathbf{n}$ terms of the series  $\mathbf{1}\times \mathbf{2}\times \mathbf{3}+\mathbf{2}\times \mathbf{3}\times \mathbf{4}+\mathbf{3}\times \mathbf{4}\times \mathbf{5}+...$

Ans:

The series provided to us is $\text{1 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 3+2 }\!\!\times\!\!\text{ 3 }\!\!\times\!\!\text{ 4+3 }\!\!\times\!\!\text{ 4 }\!\!\times\!\!\text{ 5+}...$.

Note that, the ${{n}^{th}}$ term of the given series is 

${{a}_{n}}=n\left( n+1 \right)\left( n+2 \right)$

$=\left( {{n}^{2}}+n \right)\left( n+2 \right)$

$={{n}^{3}}+3{{n}^{2}}+2n$.

Therefore, the sum to the $n$ terms of the series,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( {{k}^{3}}+3{{k}^{2}}+2k \right)} \\ & =\sum\limits_{k=1}^{n}{{{k}^{3}}}+3\sum\limits_{k=1}^{n}{{{k}^{2}}}+2\sum\limits_{k=1}^{n}{k} \\ \end{align}$

$={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{3n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{2n\left( n+1 \right)}{2}$

$={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2}+n\left( n+1 \right)$

$\begin{align} & =\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+2n+1+2 \right] \\ & =\frac{n\left( n+1 \right)}{2}\left( \frac{{{n}^{2}}+n+4n+6}{2} \right) \\ & =\frac{n\left( n+1 \right)}{2}\left( {{n}^{2}}+5n+6 \right) \\ \end{align}$

$\begin{align} & =\frac{n\left( n+1 \right)\left( {{n}^{2}}+2n+3n+6 \right)}{4} \\ & =\frac{n\left( n+1 \right)\left[ n\left( n+2 \right)+3\left( n+2 \right) \right]}{4} \\ & =\frac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{4}. \\ \end{align}$

3. Find the sum to $\mathbf{n}$ terms of the series $\mathbf{3\times }{{\mathbf{1}}^{\mathbf{2}}}\mathbf{+5\times }{{\mathbf{2}}^{\mathbf{2}}}\mathbf{+7\times }{{\mathbf{3}}^{\mathbf{2}}}\mathbf{+}...$

Ans:

The series provided to us is $3\times {{1}^{2}}+5\times {{2}^{2}}+7\times {{3}^{2}}+...$.

Note that, the ${{n}^{th}}$ term of the given series is ${{a}_{n}}=\left( 2n+1 \right){{n}^{2}}=2{{n}^{3}}+{{n}^{2}}$.

Therefore, the sum to the $n$ terms of the series is given by

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( 2{{k}^{3}}+{{k}^{2}} \right)} \\ & =2\sum\limits_{k=1}^{n}{{{k}^{3}}}+\sum\limits_{k=1}^{n}{{{k}^{2}}} \\ \end{align}$

$=2{{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

$\begin{align} & =\frac{{{n}^{2}}\left( n+1 \right)}{2}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\ & =\frac{n\left( n+1 \right)}{2}\left[ n\left( n+1 \right)+\frac{2n+1}{3} \right] \\ \end{align}$

$=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+2n+1}{3} \right]$

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

$=\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+5n+1 \right)}{6}$.

4. Find the sum to $\mathbf{n}$ terms of the series $\frac{\mathbf{1}}{\mathbf{1\times 2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2\times 3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3\times 4}}\mathbf{+}...$.

Ans:

The series provided to us, is $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$.

Note that the ${{n}^{th}}$ term of the series is ${{a}_{n}}=\frac{1}{n\left( n+1 \right)}$$=\frac{1}{n}-\frac{1}{n+1}$, using partial fractions.

Thus, for $n=1,2,3,...,n$, we have

${{a}_{1}}=\frac{1}{1}-\frac{1}{2}$

${{a}_{2}}=\frac{1}{2}-\frac{1}{3}$

${{a}_{3}}=\frac{1}{3}-\frac{1}{4}$

…

${{a}_{n}}=\frac{1}{n}-\frac{1}{n-1}$

Now, add all the terms given above. Then it gives,

${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\cdot \cdot \cdot +{{a}_{n}}=\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdot \cdot \cdot +\frac{1}{n} \right]-\left[ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot \cdot \cdot +\frac{1}{n-1} \right]$

Hence, the sum to the series up to $n$ terms is given by,

$\begin{align} & \Rightarrow {{S}_{n}}=1-\frac{1}{n+1} \\ & =\frac{n+1-1}{n+1} \\ & =\frac{n}{n+1}. \\ \end{align}$

5. Find the sum to the series ${{\mathbf{5}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{6}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{7}}^{\mathbf{2}}}\mathbf{+}...\mathbf{+2}{{\mathbf{0}}^{\mathbf{2}}}$.

Ans:

The series provided to us, is ${{\text{5}}^{\text{2}}}\text{+}{{\text{6}}^{\text{2}}}\text{+}{{\text{7}}^{\text{2}}}\text{+}...\text{+2}{{\text{0}}^{\text{2}}}$.

Note that the ${{n}^{th}}$ term of the given series is ${{a}_{n}}={{\left( n+4 \right)}^{2}}={{n}^{2}}+8n+16$.

Sum to $n$ terms of the series,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( {{k}^{2}}+8k+16 \right)} \\ & =\sum\limits_{k=1}^{n}{{{k}^{2}}}+8\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{16} \\ \end{align}$

$=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{8n\left( n+1 \right)}{2}+16n$

Now, since the ${{16}^{th}}$ term of the series, ${{\left( 16+4 \right)}^{2}}={{20}^{2}}$, so the sum of the $16$ terms,

${{S}_{n}}=\frac{16\left( 16+1 \right)\left( 2\times 16+1 \right)}{6}+\frac{8\times 16\times \left( 16+1 \right)}{2}+16\times 16$

$=\frac{\left( 16 \right)\left( 17 \right)\left( 33 \right)}{6}+\frac{8\times 16\times \left( 16+1 \right)}{2}+16\times 16$
$=\frac{16\times 17\times 33}{6}+\frac{8\times 16\times 17}{2}+256$

$\begin{align} & =1496+1088+256 \\ & =2840. \\ \end{align}$

Hence, the sum of the series is 

${{\text{5}}^{\text{2}}}\text{+}{{\text{6}}^{\text{2}}}\text{+}{{\text{7}}^{\text{2}}}\text{+}...\text{+2}{{\text{0}}^{\text{2}}}=2840$.

6. Find the sum to $\mathbf{n}$ terms of the series $\mathbf{3\times 8+6\times 11+9\times 14+}...$.

Ans:

The series provide to us, is $3\times 8+6\times 11+9\times 14+...$.

Therefore, the ${{n}^{th}}$ term of the series

$=\left( {{n}^{th}}\,\,\text{term of }\,\text{3,6,9,}... \right)\times \left( {{\text{n}}^{th}}\text{ term of 8,11,14,}... \right)$

$=3n\left( 3n+5 \right)$

$=9{{n}^{2}}+15n.$

Thus, the sum to $n$ numbers of the series,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( 9{{k}^{2}}+15k \right)} \\ & =9\sum\limits_{k=1}^{n}{{{k}^{2}}}+15\sum\limits_{k=1}^{n}{k} \\ \end{align}$

$\begin{align} & =9\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+15\times \frac{n\left( n+1 \right)}{2} \\ & =\frac{3n\left( n+1 \right)\left( 2n+1 \right)}{2}+\frac{15n\left( n+1 \right)}{2} \\ & =\frac{3n\left( n+1 \right)}{2}\left( 2n+1+5 \right) \\ & =\frac{3n\left( n+1 \right)}{2}\left( 2n+6 \right) \\ \end{align}$

$=3n\left( n+1 \right)\left( n+3 \right)$.

7. Find the sum to $\mathbf{n}$ terms of the series ${{\mathbf{1}}^{\mathbf{2}}}\mathbf{+}\left( {{\mathbf{1}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{2}}^{\mathbf{2}}} \right)\mathbf{+}\left( {{\mathbf{1}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{2}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{3}}^{\mathbf{2}}} \right)\mathbf{+}...$

Ans:

The series provided to us, is ${{1}^{2}}+\left( {{1}^{2}}+{{2}^{2}} \right)+\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)+...$.

Note that, the ${{n}^{th}}$ term of the series is,

$={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{n}^{2}}$

$=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

$\begin{align} & =\frac{n\left( 2{{n}^{2}}+3n+1 \right)}{6} \\ & =\frac{2{{n}^{3}}+3{{n}^{2}}+n}{6} \\ & =\frac{1}{3}{{n}^{3}}+\frac{1}{2}{{n}^{2}}+\frac{1}{6}n. \\ \end{align}$

Thus, the sum to $n$ terms of the series is given by

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( \frac{1}{3}{{k}^{3}}+\frac{1}{2}{{k}^{2}}+\frac{1}{6}k \right)} \\ & =\frac{1}{3}\sum\limits_{k=1}^{n}{{{k}^{3}}}+\frac{1}{2}\sum\limits_{k=1}^{n}{{{k}^{2}}}+\frac{1}{6}\sum\limits_{k=1}^{n}{k} \\ \end{align}$

$=\frac{1}{3}\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{{{2}^{2}}}+\frac{1}{2}\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{1}{6}\times \frac{n\left( n+1 \right)}{2}$

$\begin{align} & =\frac{n\left( n+1 \right)}{6}\left[ \frac{n\left( n+1 \right)}{2}+\frac{\left( 2n+1 \right)}{2}+\frac{1}{2} \right] \\ & =\frac{n\left( n+1 \right)}{6}\left[ \frac{{{n}^{2}}+n+2n+1+1}{2} \right] \\ \end{align}$

$\begin{align} & =\frac{n\left( n+1 \right)}{6}\left[ \frac{{{n}^{2}}+n+2n+2}{2} \right] \\ & =\frac{n\left( n+1 \right)}{6}\left[ \frac{n\left( n+1 \right)+2\left( n+1 \right)}{2} \right] \\ & =\frac{n\left( n+1 \right)}{6}\left[ \frac{\left( n+1 \right)\left( n+2 \right)}{2} \right] \\ \end{align}$

$=\frac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12}$.

8. Find the sum to $\mathbf{n}$ terms of the series whose ${{\mathbf{n}}^{\mathbf{th}}}$ term is $\mathbf{n}\left( \mathbf{n+1} \right)\left( \mathbf{n+4} \right)$.

Ans:

The ${{n}^{th}}$ term of the series is given as

${{a}_{n}}=\text{n}\left( \text{n+1} \right)\left( \text{n+4} \right)$.

It can be rewritten as,

${{a}_{n}}=n\left( {{n}^{2}}+5n+4 \right)={{n}^{3}}+5{{n}^{2}}+4n$.

Therefore, the sum to $n$ terms of the series,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( {{k}^{3}}+5{{k}^{2}}+4k \right)} \\ \end{align}$

$=\sum\limits_{k=1}^{n}{{{k}^{3}}}+5\sum\limits_{k=1}^{n}{{{k}^{2}}}+4\sum\limits_{k=1}^{n}{k}$

$=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+\frac{5n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{4n\left( n+1 \right)}{2}$

$\begin{align} & =\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+\frac{5\left( 2n+1 \right)}{3}+4 \right] \\ & =\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+20n+10+24}{6} \right] \\ & =\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+23n+34}{6} \right] \\ \end{align}$

$=\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+23n+34 \right)}{12}$.

9.  Find the sum to $\mathbf{n}$ terms of the series whose ${{\mathbf{n}}^{\mathbf{th}}}$ term is ${{\mathbf{n}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{2}}^{\mathbf{n}}}$.

Ans:

The ${{n}^{th}}$ term of the series is given as ${{n}^{2}}+{{2}^{n}}$.

Therefore, the sum to $n$ terms of the series,

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( {{k}^{2}}+{{2}^{k}} \right)} \\ \end{align}$

$=\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{{{2}^{k}}}$

Therefore,

${{S}_{n}}=\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{{{2}^{k}}}$                                   …… (i)

Now, note that $\sum\limits_{k=1}^{n}{{{2}^{k}}}={{2}^{1}}+{{2}^{2}}+{{2}^{3}}+\cdot \cdot \cdot +{{2}^{k}}$, is a ${{n}^{th}}$ sum of geometric progression.

Thus, $\sum\limits_{k=1}^{n}{{{2}^{k}}}=\frac{2\left( {{2}^{n}}-1 \right)}{2-1}=2\left( {{2}^{n}}-1 \right).$       …… (ii)

The equations (i) and (ii) together implies that the sum to $n$ terms of the series,

${{S}_{n}}=\sum\limits_{k=1}^{n}{{{k}^{2}}}+2\left( {{2}^{n}}-1 \right)$

$=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+2\left( {{2}^{n}}-1 \right)$.

10. Find the sum to $\mathbf{n}$ terms of the series whose ${{\mathbf{n}}^{\mathbf{th}}}$ term is ${{\left( \mathbf{2n-1} \right)}^{\mathbf{2}}}$.

Ans:

The ${{n}^{th}}$ term of the series is,

\[{{a}_{n}}={{\left( 2n-1 \right)}^{2}}=4{{n}^{2}}-4n+1\].

Thus, the sum to $n$ term of the series is given by

$\begin{align} & {{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}} \\ & =\sum\limits_{k=1}^{n}{\left( 4{{k}^{2}}-4k+1 \right)} \\ & =4\sum\limits_{k=1}^{n}{{{k}^{2}}-4\sum\limits_{k=1}^{n}{k}}+\sum\limits_{k=1}^{n}{1} \\ \end{align}$

$=\frac{4n\left( n+1 \right)\left( 2n+1 \right)}{6}-\frac{4n\left( n+1 \right)}{2}+n$

$=\frac{2n\left( n+1 \right)\left( 2n+1 \right)}{3}-2n\left( n+1 \right)+n$

$=n\left[ \frac{2\left( 2{{n}^{2}}+3n+1 \right)}{3}-2\left( n+1 \right)+1 \right]$

$=n\left[ \frac{4{{n}^{2}}+6n+2-6n-6+3}{3} \right]$

$=n\left[ \frac{4{{n}^{2}}-1}{3} \right]$

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

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 9 Exercises

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.4

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