If the sum of first n terms of an AP is cn2, then the sum of squares of these  n terms is

∵Sn=cn2∴ tn=Sn−Sn−1=c(2n−1)        Σtn2=c2Σ(2n−1)2=c2Σ4n2−4n+1=c24Σn2−4Σn+Σ1=c24n(n+1)(2n+1)6−4n(n+1)2+n=c2n232n2+3n+1−2n−2+1=c2n34n2−1=n4n2−1c23