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

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Answer : (b) 10

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 $10^3-10-990=1000-10-990=0.$

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