If the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is?

Let Sn be the sum of the first n terms of an A.P. Given that Sn = cn².
The nth term of the A.P. is given by tn = Sn - Sn-1.
Therefore, tn = cn² - c(n-1)² = cn² - c(n² - 2n + 1) = c(2n - 1).
The sum of squares of the first n terms is given by Σtn² = Σ[c(2n - 1)]² = c² Σ(4n² - 4n + 1) for n = 1 to n.
Σ(4n²) = 4Σn² = 4[n(n+1)(2n+1)/6] = 2n(n+1)(2n+1)/3
Σ(4n) = 4Σn = 4[n(n+1)/2] = 2n(n+1)
Σ(1) = n
Therefore, Σtn² = c²[2n(n+1)(2n+1)/3 - 2n(n+1) + n] = c²[2n(n+1)(2n+1) - 6n(n+1) + 3n]/3
= c²[n{2(n+1)(2n+1) - 6(n+1) + 3}]/3 = c²[n{4n² + 6n + 2 - 6n - 6 + 3}]/3
= c²[n(4n² - 1)]/3 = n(4n² - 1)c²/3
However, this seems to be incorrect based on the options provided. Let's re-examine the approach.
We have tn = c(2n - 1). Let's find the sum of squares directly:
Σtn² = Σ[c(2n-1)]² from n=1 to n = c² Σ(4n² - 4n + 1) from n=1 to n
= c² [4Σn² - 4Σn + Σ1] from n=1 to n
= c² [4(n(n+1)(2n+1)/6) - 4(n(n+1)/2) + n]
= c² [2n(n+1)(2n+1)/3 - 2n(n+1) + n]
= c²n [2(n+1)(2n+1)/3 - 2(n+1) + 1]
= c²n [ (4n²+6n+2 - 6n-6+3)/3 ]
= c²n (4n²-1)/3 = n(4n²-1)c²/3
Let's check with n=1: S1 = c, t1 = c, t1² = c², n(4n²-1)c²/3 = c² which is correct.
Let's check with n=2: S2 = 4c, t1 = c, t2 = 3c, t1² + t2² = 10c², n(4n²-1)c²/3 = 2(15)c²/3 = 10c², which is also correct.
Therefore, the sum of squares is n(4n² - 1)c²/3.