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

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

Edited By Ravindra Pindel | Updated on Jul 13, 2022 04:35 PM IST

In the previous exercises of this chapter, you have already learned about the sequences, series, progressions like arithmetic progression and geometric progression, arithmetic mean, geometric mean, etc. In the NCERT solutions for Class 11 Maths chapter 9 exercise 9.4, you will learn about some special series like the sum of first n natural numbers, the sum of the square of first n natural numbers, the sum of cubes of n natural numbers, etc.

The ways to find the sum of the series are different for different types of series, You will learn about the different ways to find the sum of the series in the NCERT book Class 11 Maths chapter 9 exercise 9.4. All the problems in exercise 9.4 Class 11 Maths give you a different methods to solve the problem. Many times problems from special series are asked in the engineering entrance exam. If you have conceptual clarity of these NCERT syllabus problems, you will be able to solve these problems easily. Check NCERT Solutions link where you will get NCERT solutions for Classes 6 to 12 at one place.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.4:-

Class 11th Maths chapter 9 exercise 9.4 consists of questions related to finding the sum of n terms of special series. There are different types of special series which are to be solved in different ways. Although Class 11 Maths chapter 9 exercise 9.4 is tougher as compared to other exercises of this chapter. But if you have solved different types of special series problems, you won't get much difficulty solving these types of questions in the exam.

Also Read| Sequences And Series Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.4:-

• NCERT solutions for Class 11 Maths chapter 9 exercise 9.4 are very important for the engineering entrance exam as many times questions from special series are asked in these exams.
• Class 11 Maths chapter 9 exercise 9.4 solutions are designed by experienced faculty who knows how to write the answer in the CBSE exam.
• You can use Class 11 Maths chapter 9 exercise 9.4 solutions for the revision of important concepts before the exam.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. What is the sum of cubes of first n natural numbers ?

Sum of cubes of first n natural numbers = [n(n+1)/2]^2

2. Find the sum of series 6+8+10+12....50

Given a = 6

d = 8-6=2

a_n= l = 50

a + (n-1)d = 50

6 + (n-1)2 = 50

n-1 = 22

n = 23

S_n = n(a+l)/2 = 23(6+50)/2 = 644

3. Find the sum of the series whose nth term is n+3

Given a_n = n+3

S_n = n(n+1)/2 + 3n

4. What is the common difference of an A.P. ?

In the arithmetic progression (A.P.), the consecutive terms of the series increase or decrease by an constant. This constant value is called the common difference of an A.P.

5. Find the arithmetic mean of two numbers 7 and 9 ?

A.M. = (7+9)/2 = 16/2 = 8

6. Find the geometric mean of two numbers 2 and 8 ?

G.M. = (2x8)^(1/2) = (16)^(1/2) = 4

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