Test Index

NCERT Class XI Mathematics - Sequences and Series - Solutions

© examsnet.com
Question : 102 of 106
Marks: +1, -0
If S1,S2,S3S_1, S_2, S_3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S229S_2^2 = S3(1+8S1)S_3(1+8S_1)
Solution:  
If 1,S2,S31_, S_2, S_3 are the sum of first n natural numbers, their squares and their cubes respectively, then
S1S_1 = ∑nk=1k\sum^{\underset{k=1}{n}} k = n(n+1)2\frac{n(n+1)}{2}
S2S_2 = ∑nk=1k2\sum^{\underset{k=1}{n}} k^2 = n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}
S3S_3 = ∑nk=1k3\sum^{\underset{k=1}{n}} k^3 = [n(n+1)2]2\left[ \frac{n(n+1)}{2} \right]^2
Now, 1 + 8S18S_1 = 1 + 8 [n(n+1)2]\left[ \frac{n(n+1)}{2} \right] = 1 + 4n (n + 1) = 4n24n^2 + 4n + 1 = (2n+1)2(2n+1)^2
Now S3S_3 (1 + 8 S1S_1) = [n(n+1)2]2(2n+1)2\left[ \frac{n(n+1)}{2} \right]^2 (2n+1)^2 = [n(n+1)(2n+1)2]2\left[ \frac{n(n+1)(2n+1)}{2} \right]^2
= 99[n(n+1)(2n+1)2]2\frac{9}{9} \left[ \frac{n(n+1)(2n+1)}{2} \right]^2 = 9[n(n+1)(2n+1)6]29\left[ \frac{n(n+1)(2n+1)}{6} \right]^2
= 9S229S_2^2
Hence, 9S229S_2^2 = S3(1+8S1)S_3(1+8S_1)
© examsnet.com
Go to Question:
Prev Question
Next Question