How Many Squares?

There are squares with side lengths $k$ where $1\leq{k}\leq100$ . With each side length $k$ , there are $(101-k)^2$ possible unit squares to be the bottom left corner of the $k\times{k}$ square, as the picture below shows.

This means there are $(101-k)^2$ squares with any side length $k$ and the total number of squares is $\sum_{k=1}^{100}(101-k)^2=100^2+99^2+...+2^2+1^2=1^2+2^2+...+99^2+100^2=\sum_{k=1}^{100}(k^2)$ .

$\sum_{k=1}^{100}(k^2)=\sum_{k=1}^{100}(\frac{k\times{2k}}{2})=\sum_{k=1}^{100}(\frac{k(2k+2)}{2}-k)=2\sum_{k=1}^{100}(\frac{k(k+1)}{2})-\sum_{k=1}^{100}(k)=\sum_{k=2}^{101}\displaystyle{k \choose 2}-\frac{100(100+1)}{2}$ .

By Hockey Stick Identity, $\sum_{k=r}^{n}\displaystyle{k\choose{r}}=\displaystyle{n+1 \choose{r+1}}$ .

We get

$\sum_{k=2}^{101}\displaystyle{k \choose 2}-\frac{100(100+1)}{2}=\displaystyle{102 \choose3}-5050=343400-5050=\boxed{338350}$ .