Let Sn​=cn2Sn−1​=c(n−1)2=cn2+c−2cnTn​=2cn−c[∵Tn​=Sn​−Sn−1​]​Tn2​=(2cn−c)2=4c2n2+c2−4c2n∴Sum=ΣTn2​=64c2⋅n(n+1)(2n+1)​+nc2−2c2n(n+1)=32c2n(n+1)(2n+1)+3nc2−6c2n(n+1)​=3nc2(4n2+6n+2+3−6n−6)​=3nc2(4n2−1)​​