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

A

$$\frac{n(4n^2 - 1)c^2}{6}$$

B

$$\frac{n(4n^2 + 1)c^2}{3}$$

C

$$\frac{n(4n^2 - 1)c^2}{3}$$

D

$$\frac{n(4n^2 + 1)c^2}{6}$$