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Binomial sum

Let m, n be positive numbers with n > m . Prove that

    \[\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{m+n-2k}{n-1} = \binom{n}{m+1}\]

Solution

Using the exercise here we have that

    \[\binom{n}{m} = \frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^n}{z^{m+1}}\, \mathrm{d}z\]

Hence,

\begin{aligned} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{m+n-2k}{n-1} &= \frac{1}{2\pi i } \sum_{k=0}^{n} (-1)^k \binom{n}{k} \oint \limits_{\left | z \right |=1} \frac{\left ( 1+z \right )^{m+n-2k}}{z^n}\, \mathrm{d}z \\ &=\frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^{m+n}}{z^n} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{\mathrm{d}z}{\left ( z+1 \right )^{2k}} \\ &= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^{m+n}}{z^n} \left ( 1 - \frac{1}{\left ( z+1 \right )^2} \right )^n \, \mathrm{d}z \\ &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\left ( z+2 \right )^n}{\left ( z+1 \right )^{n-m}}\, \mathrm{d}z \\ &=\mathfrak{Res}_{z =-1} \frac{\left ( z+2 \right )^n}{\left ( z+1 \right )^{n-m}} \\ &= \lim_{z \rightarrow -1} \frac{1}{\left ( n-m-1 \right )!} \frac{\mathrm{d}^{n-m-1} }{\mathrm{d} z^{n-m-1}} \left (\left ( z+2 \right )^n \right ) \\ &= \binom{n}{m+1} \end{aligned}

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