The correct option is C n(n+1)(2n+1)6
Here Sn=11+22+33+...n2
We consider the identity x3−(x−1)3=3x2−3x+1
When x = 1, 13−(1−1)3=3(1)2−3(1)+1
When x = 2, 23−(2−1)3=3(2)2−3(2)+1
When x = 3, 33−(3−1)3=3(3)2−3(3)+1.......
n3−(n−1)3=3n2−3n+1
Adding vertically both sides,
n3=3(12+22+32+...n2)−3(1+2+3...+n)+(1+1+1..n)
3(12+22+32+...n2)=n3+3n(n+1)2−n
= 2n3+3n2+n2
= n(2n2+3n+1)2
∑n2=n(n+1)(2n+1)6