Question

If sum of the squares of first n natural numbers exceeds their sum by 330, then find the value of n.

Answer 1

$\sum n^2$ = 330 + $\sum$n

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

⇒ $\frac{n(n+1)}{2}$ [ $\frac{2n+1}{3}$ - 1] = 330

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

⇒ n(n+1)(n-1) = 990 ⇒ n = 10.