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

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.

• If something is not true, you cannot prove that it is true. – Batominovski Dec 8 '18 at 16:46
• @Batominovski If it is wrong, then why am I asked to prove the equation. – user982787 Dec 8 '18 at 17:23
• Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences. – Batominovski Dec 8 '18 at 17:26

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

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

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

\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}

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.