My first Combinatorics problem!

Note in a 1 by 1, there are 1 squarea, and 2 by 2, there are 5 squares, 3 by 3 is 9. Thus in n by n we get:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.$

As required for n=8, we get:

$\frac{8(8+1)(2\cdot8+1)}{6} \\=\frac{8\cdot 9\cdot 17}{6}=\frac{2\cdot 4 \cdot 3 \cdot 3 \cdot 17}{2\cdot 3}= \\ 4\cdot 3\cdot 17 = 4 \cdot 51 = 204$