Question

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

Answer 1

the answer is 10 sum of first natural numbers by formula

Answer 2

Answer:

10

Step-by-step explanation:

Given : Sum of squares of first n natural numbers exceeds their sum by 330,

To Find : n

Solution:

Sum of squares of first n natural =$\frac{n(n+1)(2n+1)}{6}$

Sum of natural numbers = $\frac{n(n+1)}{2}$

Since we are given that Sum of squares of first n natural numbers exceeds their sum by 330.

A.T.Q

$\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=330$

$\frac{n(n+1)(2n+1)-3n(n+1)}{6}=330$

$n(n+1)(2n+1)-3n(n+1)=330 \times 6$

$2n^3-2n^2-1980=0$

$n=10,-5-i\sqrt{74},-5+i\sqrt{74}$

Since n is the number of terms

So, n = 10

Hence the value of n is 10