Squares Counting..

we have the formula : $\frac{n\times(n+1)\times(2n+1)}{6}$ for number of squares in a chessboard of any size.

So, substituting the value n=5 for an 5 X 5 chessboard.

$\Rightarrow No. \ of \ squares = \frac{n\times(n+1)\times(2n+1)}{6}$ $= \frac{5 (5+1) (10+1)}{6}$ $= \frac{330}{6}$ $= \boxed{ 55}$ .

the formula is the sum of squares from1 to 5 i.e. $1^{2} +2^{2} +3^{2} +4^{2}+5^{2}$ $\boxed{=55}$

An easier solution exist where size of n is smaller : (n^2)+((n-1)^2)+((n-2)^2)+...+1. In this case , it would be , 5^2 + 4^2 + 3^2 + 2^2 + 1^2 . So ,it would be 25 + 16 + 9 + 4 + 1 .