If sum of first n terms of an A.P. is c n^2 , then sum of squares of these n terms is

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8. Sequences and Series

hard

If the sum of first $n$ terms of an $A.P.$ is $c n^2$, then the sum of squares of these $n$ terms is

A

$\frac{n\left(4 n^2-1\right) c^2}{6}$

B

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

C

$\frac{n\left(4 n^2-1\right) c^2}{3}$

D

$\frac{n\left(4 n^2+1\right) c^2}{6}$

(IIT-2009)

Solution

$S _{ n }= cn ^2$

$S_{ n -1}= c ( n -1)^2= cn ^2+ c -2 cn$

$T _{ n }=2 cn – c$

$T _{ n }^2=(2 cn – c )^2=4 c ^2 n ^2+ c ^2-4 c ^2 n$

Required sum

$=\sum T_n^2=4 c^2 \sum n^2+n c^2-4 c^2 \sum n$

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

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

$=\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3}$

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

Standard 11

Mathematics

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