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Given $\sum_{k=0}^{2r} (-1)^k \binom{n}{k}\binom{n}{2r-k} = (-1)^r\binom{n}{r}$ for $0≤r≤\frac12n$

How do I show that $\sum_{k=0}^{r} (-1)^k \binom{n}{k}\binom{n}{2r-k} = \frac12(-1)^r\binom{n}{r}[1+\binom{n}{r}]$ for $0≤r≤\frac12n$

For context, the first statement is derived from considering the coefficient of $x^{2r}$ in the statement $(1-x)^n(1+x)^n=(1-x^2)^n$ where $2r≤n$

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Use symmetry: we split the sum into two parts and then we change the index in the second one by letting $j=2r-k$, $$\begin{align} \sum_{k=0}^{2r} (-1)^k \binom{n}{k}&\binom{n}{2r-k}=\sum_{k=0}^{r} (-1)^k \binom{n}{k}\binom{n}{2r-k}+\sum_{k=r+1}^{2r} (-1)^k \binom{n}{k}\binom{n}{2r-k}\\ &=\sum_{k=0}^{r} (-1)^k \binom{n}{k}\binom{n}{2r-k}+\sum_{j=0}^{r-1} (-1)^{2r-j} \binom{n}{2r-j}\binom{n}{j}\\ &=\sum_{k=0}^{r} (-1)^k \binom{n}{k}\binom{n}{2r-k}+\sum_{j=0}^{r} (-1)^{j} \binom{n}{2r-j}\binom{n}{j}- (-1)^{r}\binom{n}{r}^2\\ &=2\sum_{k=0}^{r} (-1)^k \binom{n}{k}\binom{n}{2r-k}- (-1)^{r}\binom{n}{r}^2. \end{align}$$

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  • $\begingroup$ I'm having a bit of trouble understanding how exactly the second line follows from the first. Are you using $\binom{n}{r}=\binom{n}{n-r}$ to get the result? If it's not too much to ask, could you show an extra line of working between the two? $\endgroup$ Jul 21, 2021 at 8:31
  • $\begingroup$ @Quippy I edited my answer with an extra line. Is it better now? $\endgroup$
    – Robert Z
    Jul 21, 2021 at 9:08
  • $\begingroup$ I'm so sorry I meant I don't get how $\sum_{k=r+1}^{2r} (-1)^k \binom{n}{k}\binom{n}{2r-k}=$$\sum_{j=0}^{r-1} (-1)^{2r-j} \binom{n}{2r-j}\binom{n}{j}$ $\endgroup$ Jul 21, 2021 at 9:32
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    $\begingroup$ @Quippy Change index of summation: let $j=2r-k$ then $k=r+1,\dots, 2r$ becomes $j=0,\dots, r-1$. $\endgroup$
    – Robert Z
    Jul 21, 2021 at 9:35
  • $\begingroup$ Thanks for taking the time to help :) $\endgroup$ Jul 21, 2021 at 9:39

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