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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 =

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

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

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

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

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

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

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

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

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

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

Hence, the sum is

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

Question:5 Find the sum to n terms of each of the series in $5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2$

Answer:

series = $5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2$

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 $(16+4)^2=20^2$

$S_1_6=\frac{16(16+1)(2(16)+1)}{6}+\frac{8.(16)(16+1)}{2}+16(16)$

$S_1_6=\frac{16(17)(33)}{6}+\frac{8.(16)(17)}{2}+16(16)$

$S_1_6=1496+1088+256$

$S_1_6=2840$

Hence, the sum of the series $5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2$ 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 = $3n(3n+5)=a_n=9n^2+15n$

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

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

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

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

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

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

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

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

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

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

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 $n (n+1) ( n + 4 )$

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

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

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

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

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

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

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

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

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

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 $( 2n-1) ^2$.

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

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

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

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

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

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

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

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

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