Answer: $\mathbf C$
Explanation:
$\mathrm {S_n = cn^2} \text{[Given]}$
Now $\mathrm {n^{th}}$ term from the sum can be calculated by using, the formula:
$\mathrm {\therefore t_n = S_n - S_{n-1} = cn^2-c(n-1)^2 = c(2n-1)}$
Now,
$\mathrm {\sum t_{n^2} = \sum c^2(2n-1)^2 = \sum c^2(4n^2+1-4n)}\tag{i}$
Now, we know that:
$\mathrm {\sum n^2 = \frac{n(n+1)(2n+1)}{6}\;,\; \sum n = \frac{n(n+1)}{2}\;, and\;\sum 1 = n}$
Substituting these values in $\mathrm {(i)}$, we get:
$\mathrm {\sum t_{n^2} = c^2\Big(4\frac{n(n+1)(2n+1)}{6}\;+n + \frac{n(n+1)}{2}\Big) = \frac{c^2(4n^3-n)}{3} = \frac{n(4n^2-1)c^2}{3}}$
$\therefore \mathbf C$ is the right option.