Alternating sum of a part of a row of Pascal's triangle

Question

$\displaystyle \sum_{r=0}^m(-1)^r{n \choose r}=(-1)^m{n-1 \choose m}$ if $m$ is less than $n$.

This question actually consists of two part that is when $m$ is less than $n$ and when $m$ is equal to $n$. I can solve the second part but not the first part.

Answer 1

Hint: Use induction on $m$. This boils down to proving $$ (-1)^m \binom{n-1}{m} + (-1)^{m+1} \binom{n}{m+1} = (-1)^{m+1} \binom{n-1}{m+1}, $$ which follows from Pascal's identity.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r} =\pars{-1}^{m}{n - 1 \choose m}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r}} =\sum_{r\ =\ 0}^{m}\pars{-1}^{r}\ \overbrace{\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{n} \over z^{r + 1}}\,{\dd z \over 2\pi\ic}}^{\ds{=\ {n \choose r}}} \\[5mm]&=\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{n} \over z}\sum_{r\ =\ 0}^{m}\pars{-\,{1 \over z}}^{r} \,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{n} \over z}{\pars{-1/z}^{m + 1} - 1 \over -1/z - 1} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{n} \over z}\,{z^{m + 1} + \pars{-1}^{m} \over z^{m}\pars{1 + z}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\ \underbrace{\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n - 1} \,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}}\ +\ \pars{-1}^{m}\ \underbrace{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z^{m + 1}} \,{\dd z \over 2\pi\ic}}_{\ds{=\ {n - 1 \choose m}}} \end{align}

$$\color{#66f}{\large\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r} =\pars{-1}^{m}{n - 1 \choose m}} $$