TITLE: How do I evaluate $\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$?
QUESTION [7 upvotes]: $$\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$$

This is what Wolfram says:

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

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}{n^{2} + n \choose r} & =
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}{n^{2} + n \choose n^{2} + n - r} =
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}
\bracks{z^{n^{2} + n - r}}\pars{1 + z}^{n^{2} + n}
\\[5mm] & =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n}\sum_{r = 0}^{n^{2}}\pars{-z}^{r} =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n}
\,{\pars{-z}^{n^{2} + 1} - 1 \over -z - 1}
\\[5mm] & =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n - 1} +
\pars{-1}^{n^{2}}\bracks{z^{n - 1}}\pars{1 + z}^{n^{2} + n - 1}
\\[5mm] & =\
\underbrace{{n^{2} + n - 1 \choose n^{2} + n}}_{\ds{=\ 0}}\ +\ \pars{-1}^{n^{2}}{n^{2} + n - 1 \choose n - 1} =
\bbx{\pars{-1}^{n^{2}}{n^{2} + n - 1 \choose n - 1}}
\end{align}