Sum of squares of first n natural numbers exceeds their sum by 330, then n =

$(a)\;8\qquad(b)\;10\qquad(c)\;15\qquad(d)\;20$

Explanation : sum of sqaures of first natural numbers $\;\frac{n}{6(n+1)(2n+1)}$
Sum of first n natural numbers $\frac{n(n+1)}{2}$
Given that $\;\frac{n}{6(n+1)(2n+1)}=\frac{n(n+1)}{2}+330$
Therefore , $\frac{n}{6(n+1)[2n+1-3]}-330=0$
$\frac{n}{3(n+1)(n-1)}-330=0$
$n^3-n-990=0$
Substituting each option we get $10^3-10-990=1000-10-990=0.$