Proving $\binom{n+1+m}{n+1}=\sum_{k=0}^m(k+1)\binom{n+m-k}{n}$

Question

Use any method to prove that $$\binom{n+1+m}{n+1}=\sum_{k=0}^m(k+1)\binom{n+m-k}{n}$$

My Try:

Base case: Let $m=1$

LHS$$\binom{n+1+m}{n+1}=\binom{n+2}{n+1}=(n+2)$$ RHS$$\sum_{k=0}^m(k+1)\binom{n+m-k}{n}=\binom{n+1-0}{n}+(1+1)\binom{n+1-1}{n}$$ $$=\frac{(n+1)!}{n!}+2$$ $$=(n+3)$$

If $m=2$

LHS$$\binom{n+3}{n+1}=\frac{n^2+5n+6}{2!}$$ RHS$$=\binom{n+2}{n}+2\binom{n+1}{n}+3\binom{n}{n}$$ $$=\frac{n^2+3n+2+4n+4+6}{2}=\frac{n^2+7n+12}{2}$$

Clearly $LHS\ne RHS$

If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\require{cancel}\bcancel{\cancel{n + 1 + m \choose n + 1}} = \sum_{k = 0}^{m}\pars{k + 1}{n + m - k \choose n}:\ {\LARGE ?}}$.

The right answer is $\bbx{\ds{n + m + 2 \choose n + 2}}$

$$ \bbx{\mbox{Note that}\ {n + m - k \choose n} = 0\ \mbox{when}\ k > m} $$

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\begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{m}\pars{k + 1} {n + m - k \choose n}} = \sum_{k = 0}^{\infty}\pars{k + 1}{n + m - k \choose m - k} \\[5mm] = &\ \sum_{k = 0}^{\infty}\pars{k + 1} \bracks{{-n - 1 \choose m - k}\pars{-1}^{m - k}} \\[5mm] = &\ \pars{-1}^{m}\sum_{k = 0}^{\infty}\pars{k + 1}\pars{-1}^{k} \bracks{z^{m - k}}\pars{1 + z}^{-n - 1} \\[5mm] = &\ \pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1} \sum_{k = 0}^{\infty}\pars{k + 1}\pars{-z}^{k} \\[5mm] = &\ \pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1}\, \pars{-\,\partiald{}{z}\sum_{k = 0}^{\infty}\pars{-z}^{k + 1}} \\[5mm] = &\ \pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1}\, \partiald{}{z}\pars{z \over 1 + z} = \pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 3} \\[5mm] = &\ \pars{-1}^{m}{-n - 3 \choose m} = \pars{-1}^{m}\bracks{{n + 3 + m - 1\choose m}\pars{-1}^{m}} \\[5mm] = &\ \bbx{n + m + 2 \choose n + 2} \end{align}

Answer 2

The RHS can be written as $$\sum_{i+j=n+m+1}\binom{i}1\binom{j}n$$where $\binom{r}{s}:=0$ if $s\notin\{0,\dots,r\}$.

This equals: $$\binom{n+2+m}{n+2}$$ See here for a proof of that. So RHS does not equal LHS.