Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

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


Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App