Question

The sum of squares of first n natural numbers is given by $\frac{1}{6}n(n+1)(2n+1)$ or $\frac{1}{6}(2n^3+3n^2+n)$. Find the sum of squares of the first 10 natural numbers.

Answer 1

Given, the sum of squares of first n natural numbers $=\frac{1}{6}n(n+1)(2n+1)$
$\therefore$ The sum of squares of first 10 natural numbers
$=\frac{1}{6}\left(10\right)(10+1)(2\times 10+1)=\frac{1}{6}\times 10\times 11\times 21$ [put n = 10]
= 385