Question

What is the sum of first N natural number's squares?

Answer 1

you mean
 $1^2+ 2^2+3^2 +... n^2 = ..?$ 

Then its equal to
$\frac{n(n+1)(2n+1)}{6}$

Answer 2

The sum of the squares of the first 'n' natural numbers

1² + 2² + 3² + 4² + 5² ..............

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