Q23 / 57 · JEE Advanced 2009 (Paper 2) · Mathematics
If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$ terms is
- A$\frac{n\left(4 n^2-1\right) c^2}{6}$
- B$\frac{n\left(4 n^2+1\right) c^2}{3}$
- Correct answerC$\frac{n\left(4 n^2-1\right) c^2}{3}$
- D$\frac{n\left(4 n^2+1\right) c^2}{6}$
Solution
Let $S_n=c^2$
$$
\begin{aligned}
& S_{n-1}=c(n-1)^2=c n^2+c-2 c n \\
& \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \\
& T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \\
& \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \\
& =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \\
& =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \\
& =\frac{n c^2\left(4 n^2-1\right)}{3}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.