2017/011) If the sum of first n terms of an A.P. is $cn^2$ find the sum of squares of these n terms.

the sum of $1^{st}$ n terms is $cn^2$
so $k^{th}$ term is $ck^2-c(k-1)^2= c*(2k-1)$
so sum of n terms = $\sum_{k=1}^{n} (c^2(2k-1)^2) = c^2 \sum_{k=1}^{n} (4k^2-4k + 1) = c^2(4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2} +n)$
= $\frac{nc^2(4n^2-1)}{3}$