How many squares do you see?

$\text{solution}^{1}$

$\text{1+4+9+16+25+36+49+64+81+100=385}$ .

$\text{solution}^{2}$ .

$\quad \sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

$1^2 + 2^2 + 3^2 + \ldots + 10^2 = \dfrac{10(11)(21)}{6} = \boxed{385}$

The number of squares is $1^2 + \ldots + n^2$ , for a $n \times n$ square.

Putting $n = 10$

$1^2 + 2^2 + 3^2 + \ldots + 10^2 = \dfrac{10(11)(21)}{6} = \boxed{385}$

It would have been better for the solvers to have figured out the pattern themselves. You made the problem too easy: Level 0.