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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
