In the sum of first $n$ terms of an A.P. is $cn^{2}$, then the sum of squares of these $n$ terms is

In the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is $\text{Given: } S_n = cn^{2}$ $\text{To find: } a_1^{2} + a_2^{2} + \cdots + a_n^{2}$ $1.\ \text{find } a_n \text{ from } S_n - S_{n-1}$ $2.\ \text{identify the A.P. } (a,\ d)$ $3.\ \text{sum the squares and match the option}$ $S_{n-1} = c\,(n-1)^{2}$ $a_n = S_n - S_{n-1}$ $a_n = cn^{2} - c(n-1)^{2}$ $a_n = c\left[n^{2} - (n-1)^{2}\right]$ $(n-1)^{2} = n^{2} - 2n + 1$ $a_n = c\left(n^{2} - n^{2} + 2n - 1\right) = c(2n-1)$ $a = c, \quad d = 2c$ $a_k = c\,(2k-1), \quad k = 1, 2, \ldots, n$ $a_1^{2} + a_2^{2} + \cdots + a_n^{2} = \sum_{k=1}^{n} c^{2}(2k-1)^{2}$ $= c^{2}\sum_{k=1}^{n}\left(4k^{2} - 4k + 1\right)$ $(2k-1)^{2} = 4k^{2} - 4k + 1$ $= c^{2}\left(4\sum_{k=1}^{n} k^{2} - 4\sum_{k=1}^{n} k + n\right)$ $\sum_{k=1}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}$ $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ $= c^{2}\left[\dfrac{4n(n+1)(2n+1)}{6} - \dfrac{4n(n+1)}{2} + n\right]$ $= c^{2}\left[\dfrac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n\right]$ $= c^{2}\,n\left[\dfrac{2(n+1)(2n+1) - 6(n+1) + 3}{3}\right]$ $2(n+1)(2n+1) = 4n^{2} + 6n + 2$ $4n^{2} + 6n + 2 - 6n - 3 = 4n^{2} - 1$ $a_1^{2} + a_2^{2} + \cdots + a_n^{2} = c^{2}\,\dfrac{n(4n^{2}-1)}{3} = \dfrac{n(4n^{2}-1)c^{2}}{3}$ $\boxed{\text{Answer: (3)}\ \dfrac{n(4n^{2}-1)c^{2}}{3}}$