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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Sum of squares of first n natural numbers exceeds their sum by 330, then n =


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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 , $\frac{n}{6(n+1)[2n+1-3]}-330=0$
Substituting each option we get $10^3-10-990=1000-10-990=0.$
answered Jan 17, 2014 by yamini.v

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