How do I evaluate $\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$?

Question

$$\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$$

This is what Wolfram says:

Is there any simple way to prove it? Like with counting arguments or manipulating binomial expansions?

Answer 1

This is a case of $$\sum_{r=0}^k(-1)^r\binom mr=(-1)^k\binom{m-1}k$$ which is easily proved by induction on $k$.

Answer 2

Using Vandermonde's Identity: $$ \begin{align} \sum_{r=0}^{n^2}(-1)^r\binom{n^2+n}{r} &=\sum_{r=0}^{n^2}(-1)^{n^2}\binom{n^2+n}{r}\binom{-1}{n^2-r}\\ &=(-1)^{n^2}\binom{n^2+n-1}{n^2} \end{align} $$