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

Sequences And Series Class 11 Chapter 9 Exercise: 9.4

Question:1 Find the sum to n terms of each of the series in $1\times 2+2\times 3+3\times 4+4\times 5+...$

Answer:

the series = $1\times 2+2\times 3+3\times 4+4\times 5+...$

n th term = $n(n+1)=a_n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^{n}k(k+1)$

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

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

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

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

$=\frac{n(n+1)}{2}\left(\frac{(2n+4)}{3}\right)$

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

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

Question:2 Find the sum to n terms of each of the series in $1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...$

Answer:

the series = $1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...$

n th term = $n(n+1)(n+2)=a_n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^{n}k(k+1)(k+2)$

$=\sum _{k=1}^n{k}^3+3\sum _{k=1}^n{k}^2+2\sum _{k=1}^{n}k$

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

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

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

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

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

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

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

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

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

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

Thus, sum is

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

Question:3 Find the sum to n terms of each of the series $3\times 1^2+5\times 2^2+7\times +....+{20}^2$

Answer:

the series $3\times 1^2+5\times 2^2+7\times +....+{20}^2$

nth term = $(2n+1)(n^2)=2n^3+n^2=a_n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^{n}2k^3+k^2$

$=2\sum _{k=1}^n{k}^3+\sum _{k=1}^n{k}^2$

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

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

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

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

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

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

Thus, the sum is

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

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

Answer:

Series =

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$

$n^{th}term=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$a_1=\frac{1}{1}-\frac{1}{2}$

$a_2=\frac{1}{2}-\frac{1}{3}$

$a_3=\frac{1}{3}-\frac{1}{4}$.................................

$a_n=\frac{1}{n}-\frac{1}{n+1}$

$a_1+a_2+a_3+...................a_n=[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+............\frac{1}{n}]-[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.............\frac{1}{n+1}]$

$a_1+a_2+a_3+...................a_n=\left[\frac{1}{1}\right]-\left[\frac{1}{n+1}\right]$

$S_n=\frac{n+1-1}{n+1}$

$S_n=\frac{n}{n+1}$

Hence, the sum is

$S_n=\frac{n}{n+1}$

Question:5 Find the sum to n terms of each of the series in $5^2+6^2+7^2+....+20^2$

Answer:

series = $5^2+6^2+7^2+....+20^2$

n th term = $(n+4{)}^2=n^2+8n+16=a_n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n(k+4{)}^2$

$=\sum _{k=1}^n{k}^2+8\sum _{k=1}^{n}k+\sum _{k=1}^{n}16$

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

16th term is $(16+4{)}^2={20}^2$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

Hence, the sum of the series $5^2+6^2+7^2+....+20^2$ is 2840.

Question:6 Find the sum to n terms of each of the series $3\times 8+6\times 11+9\times 14+...$

Answer:

series = $3\times 8+6\times 11+9\times 14+...$

=(n th term of 3,6,9,...........)$\times$(nth terms of 8,11,14,..........)

n th term = $3n(3n+5)=a_n=9n^2+15n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^{n}3k(3k+5)$

$=9\sum _{k=1}^n{k}^2+15\sum _{k=1}^{n}k$

$=\frac{9.n(n+1)(2n+1)}{6}+\frac{15.n(n+1)}{2}$

$=\frac{3.n(n+1)(2n+1)}{2}+\frac{15.n(n+1)}{2}$

$=\frac{n(n+1)}{2}(3(2n+1)+15)$

$=\frac{3.n(n+1)}{2}(2n+1+5)$

$=\frac{3.n(n+1)}{2}(2n+6)$

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

$=3.n(n+1)(n+3)$

Hence, sum is $=3.n(n+1)(n+3)$

Question:7 Find the sum to n terms of each of the series in $1^2+(1^2+2^2)+(1^2+2^2+3^2)...$

Answer:

series = $1^2+(1^2+2^2)+(1^2+2^2+3^2)...$

n th term = $a_n=1^2+2^2+3^2+...................n^2=\frac{n(n+1)(2n+1)}{6}$

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

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n\frac{2k^3+3k^2+k}{6}$

$=\frac{1}{3}\sum _{k=1}^n{k}^3+\frac{1}{2}\sum _{k=1}^n{k}^2+\frac{1}{6}\sum _{k=1}^{n}k$

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

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

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

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

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

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

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

Question:8 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by $n(n+1)(n+4)$

Answer:

nth terms is given by $n(n+1)(n+4)$

$a_n=n(n+1)(n+4)=n(n^2+5n+4)=n^3+5n^2+4n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^{n}k(k+1)(k+4)$

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

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

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

$=\left[\frac{n(n+1)}{2}\right](\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)$

$=\left[\frac{n(n+1)}{2}\right]\left(\frac{3n^2+3n+20n+10+24}{6}\right)$

$=\left[\frac{n(n+1)}{2}\right]\left(\frac{3n^2+23n+34}{6}\right)$

$=\left[\frac{n(n+1)}{24}\right](3n^2+23n+34)$

Question:9 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by $n^2+2^n$

Answer:

nth terms are given by $n^2+2^n$

$a_n=n^2+2^n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n{k}^2+\sum _{k=1}^n{2}^k$

$\sum _{k=1}^n{2}^k=2^1+2^2+2^3+....................{.2}^n$

This term is a GP with first term =a =2 and common ratio =r =2.

$\sum _{k=1}^n{2}^k$$=\frac{2(2^n-1)}{2-1}=2(2^n-1)$

$S_n=\sum _{k=1}^n{k}^2+2(2^n-1)$

$S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)$

Thus, the sum is

$S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)$

Question:10 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by $(2n-1{)}^2$

Answer:

nth terms is given by $(2n-1{)}^2$.

$a_n=(2n-1{)}^2=4n^2+1-4n$

$S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n(2k-1{)}^2$

$=4\sum _{k=1}^n{k}^2-4\sum _{k=1}^{n}k+\sum _{k=1}^{n}1$

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

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

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

$=n(\frac{4n^2+6n+2-6n-6+3}{3})$

$=n(\frac{4n^2-1}{3})$

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