Question

If the sum of first $n$ terms of an AP is $cn^2$, then the sum of squares of these $n$ terms is

• A. $\frac{n\, (4n^2-1)\, c^2}{6}$
• B. $\frac{n\, (4n^2+1)\, c^2}{3}$
• C. $\frac{n\, (4n^2-1)\, c^2}{3}$
• D. $\frac{n\, (4n^2+1)\, c^2}{6}$

Answer 1

Answer: (C)

Let $s_n = cn^2$
$ S_{n-1} = c \, (n-1)^2 = cn^2 \, + c \, -2cn$
$\therefore T_n = 2cn -c [\, \because \, T_n = S_n\, - S_{n-1}]$
$ T^2_n = (2cn -c)^2 = 4c^2n^2 +c^2 - 4c^2n $
$\therefore $ Sum $= \displaystyle \sum T^2_n = \frac{4c^2.n (n+1)\, (2n+1)}{6} + nc^2 - 2c^2n\, (n+1)$
$ =\frac{2c^2n \, (n+1)\, (2n+1)+ 3nc^2 -6c^2n\, (n+1)}{3}$
$ =\frac{nc^2(4n^2 \, +6n \, +2 \, +3 \, -6n \, -6) }{3} = \frac{nc^2\, (4n^2-1)}{3}$