Show that $ \sum_k \binom{r}{k} \binom{s+k}{n} (-1)^k = (-1)^r \binom{s}{n-r} $

Question

I can't resolve this exercise and I need tips.

Let be $n$ integer, $s$ real and $r \geq 0$ integer. Show that

$$ \sum_k \binom{r}{k} \binom{s+k}{n} (-1)^k = (-1)^r \binom{s}{n-r} $$

Answer 1

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

\begin{align} &\color{#66f}{\large\sum_{k}{r \choose k}{s + k \choose n}\pars{-1}^{k}} =\sum_{k}{r \choose k}\pars{-1}^{k}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s + k} \over z^{n + 1}} \,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n + 1}} \sum_{k}{r \choose k}\pars{-1 - z}^{k}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n + 1}} \bracks{1 + \pars{-1 - z}}^{r}\,{\dd z \over 2\pi\ic} \\[3mm]&=\pars{-1}^{r}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n - r + 1}} \,{\dd z \over 2\pi\ic} =\color{#66f}{\large\pars{-1}^{r}{s \choose n - r}} \end{align}

Answer 2

Vandermonde's Identity says $$ \binom{s+k}{n}=\sum_{j=0}^n\binom{s}{n-j}\binom{k}{j}\tag{1} $$ Thus, $$ \begin{align} \sum_{k=0}^r(-1)^k\binom{r}{k}\binom{s+k}{n} &=\sum_{k=0}^r(-1)^k\binom{r}{k}\sum_{j=0}^n\binom{s}{n-j}\binom{k}{j}\tag{2}\\ &=\sum_{k=0}^r\sum_{j=0}^n(-1)^k\binom{r}{k}\binom{k}{j}\binom{s}{n-j}\tag{3}\\ &=\sum_{k=0}^r\sum_{j=0}^n(-1)^k\binom{r}{j}\binom{r-j}{k-j}\binom{s}{n-j}\tag{4}\\ &=(-1)^r\binom{s}{n-r}\tag{5} \end{align} $$ Explanation:
$(2)$: apply $(1)$
$(3)$: rearrange terms
$(4)$: $\binom{r}{k\vphantom{j}}\binom{k}{j}=\binom{r}{j}\binom{r-j}{k-j}$ (just write out the coefficients in terms of factorials)
$(5)$: $\sum\limits_{k=0}^r(-1)^k\binom{r-j}{k-j}=\sum\limits_{k=0}^{r-j}(-1)^{k+j}\binom{r-j}{k}=(-1)^j(1-1)^{r-j}=(-1)^j[j=r]$

Answer 3

I'm just posting this to add what I did starting from $(4)$ in robjohn answer.

$$ \sum_{k=0}^r(-1)^k\binom{r}{k}\binom{s+k}{n}=\sum_{k=0}^r\sum_{j=0}^n(-1)^k\binom{r}{j}\binom{r-j}{k-j}\binom{s}{n-j} $$ In order to keep the inner sum consistent, we have to restrict the values of $j$ to $j=0,1,2,\cdots,r$ so this way the $\binom{r}{j}$ is always non zero, then:

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

Now setting $j=t$ a fixed value of $j$ on the inner sum, which can be $0,1,2,\cdots,r$, we can examine what happens to the overall sum. So we have

$$ \sum_{k=0}^r(-1)^k\binom{r}{t}\binom{r-t}{k-t}\binom{s}{n-t}=\binom{r}{t}\binom{s}{n-t}\sum_{k=0}^r(-1)^k\binom{r-t}{k-t} \tag{*} $$

Here, at right hand side we see that the factor $\binom{r-t}{k-t}$ is non zero only if $k \geq t$, then:

$$ \sum_{k=0}^r(-1)^k\binom{r}{t}\binom{r-t}{k-t}\binom{s}{n-t}=\binom{r}{t}\binom{s}{n-t}\sum_{k=t}^r(-1)^k\binom{r-t}{k-t} $$

Shifting index in the right hand side sum:

$$ \sum_{k=0}^r(-1)^k\binom{r}{t}\binom{r-t}{k-t}\binom{s}{n-t}=\binom{r}{t}\binom{s}{n-t}\sum_{k=0}^{r-t}(-1)^{k+t}\binom{r-t}{k} $$ $$ =(-1)^t\binom{r}{t}\binom{s}{n-t} \sum_{k=0}^{r-t}\binom{r-t}{k}(-1)^{k} =(-1)^t\binom{r}{t}\binom{s}{n-t}(1-1)^{r-t} =0 $$ For all $t<r$.

But for $t=r$ , from $($*$)$ we have that:

$$ \sum_{k=0}^r(-1)^k\binom{r}{r}\binom{r-r}{k-r}\binom{s}{n-r}=\binom{s}{n-r}\sum_{k=0}^{r}(-1)^{k}\binom{0}{k-r} $$

Where the facotr $\binom{0}{k-r}$ is non zero only when $k=r$. Therefore, the only term in $($*$)$ that remains is the one with $j=k=r$, which is: $$ \binom{s}{n-r}(-1)^{r}\binom{0}{r-r}=\binom{s}{n-r}(-1)^{r}\binom{0}{0}=(-1)^{r}\binom{s}{n-r} $$

Finally:

$$ \sum_{k=0}^r\sum_{j=0}^r(-1)^k\binom{r}{j}\binom{r-j}{k-j}\binom{s}{n-j}=(-1)^{r}\binom{s}{n-r} $$