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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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Answer:
the series = $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ...$
n th term = $n(n+1)=a_n$
$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)$
$=\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} k$
$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{2}\left ( \frac{(2n+1)}{3}+1 \right )$
$=\frac{n(n+1)}{2}\left ( \frac{(2n+1+3)}{3} \right )$
$=\frac{n(n+1)}{2}\left ( \frac{(2n+4)}{3} \right )$
$=n(n+1)\left ( \frac{(n+2)}{3} \right )$
$= \frac{n(n+1)(n+2)}{3}$
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
$=\left [ \frac{n(n+1)(n+2)(n+3)}{4} \right ]$
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
$= \frac{n(n+1)(3n^2+5n+1)}{6}$
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 $=3.n(n+1)\left (n+3\right )$
Answer:
series = $1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...$
n th term = $a_n=1^2+2^2+3^2+...................n^2=\frac{n(n+1)(2n+1)}{6}$
$=\frac{n(2n^2+3n+1)}{6}=\frac{2n^3+3n^2+n}{6}$
$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{2k^3+3k^2+k}{6}$
$=\frac{1}{3}\sum _{k=1}^{n} k^3+\frac{1}{2}\sum _{k=1}^{n} k^2+\frac{1}{6}\sum _{k=1}^{n} k$
$=\frac{1}{3}\left [ \frac{n(n+1)}{2} \right ]^2+\frac{1}{2}.\frac{n(n+1)(2n+1)}{6}+\frac{1}{6}\frac{n(n+1)}{2}$
$=\left [ \frac{n(n+1)}{6} \right ] (\frac{n(n+1)}{2}+\frac{2n+1}{2}+\frac{1}{2})$
$=\left [ \frac{n(n+1)}{6} \right ] (\frac{n^2+n+2n+1+1}{2})$
$=\left [ \frac{n(n+1)}{6} \right ] (\frac{n^2+n+2n+2}{2})$
$=\left [ \frac{n(n+1)}{6} \right ] (\frac{n(n+1)+2(n+1)}{2})$
$=\left [ \frac{n(n+1)}{6} \right ] (\frac{(n+1)(n+2)}{2})$
$=\left [ \frac{n(n+1)^2(n+2)}{12} \right ]$
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 $( 2n-1) ^2$.
$a_n=( 2n-1) ^2=4n^2+1-4n$
$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (2k-1)^2$
$=4\sum _{k=1}^{n} k^2-4\sum _{k=1}^{n} k+\sum _{k=1}^{n} 1$
$=\frac{4.n(n+1)(2n+1)}{6}-\frac{4.n(n+1)}{2}+n$
$=\frac{2.n(n+1)(2n+1)}{3}-2.n(n+1)+n$
$=n[\frac{2(n+1)(2n+1)}{3}-2(n+1)+1]$
$=n(\frac{4n^2+6n+2-6n-6+3}{3})$
$=n(\frac{4n^2-1}{3})$
$=n(\frac{(2n+1)(2n-1)}{3})$
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
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Happy learning!!!
Sum of cubes of first n natural numbers = [n(n+1)/2]^2
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
Given a_n = n+3
S_n = n(n+1)/2 + 3n
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.
A.M. = (7+9)/2 = 16/2 = 8
G.M. = (2x8)^(1/2) = (16)^(1/2) = 4
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