Finding $\sum_{r=0}^{2n+1}(-1)^{r}{2n+1\choose r}^2$ I have to find the value of the following sum involving binomial coefficients. According to what I have solved, the answer should be $0$ but as per the answers given it is $1$. It would be great if someone could identify the error.

$$\sum_{r=0}^{2n+1}(-1)^{r}{2n+1\choose r}^2$$


Consider the following expansions. The required summation is then just the coefficient of $x^{2n+1}$ in $(1+x)^{2n+1}[-(x-1)^{2n+1}]$ as can be seen from the following.
$$\begin{aligned}(1+x)^{2n+1}&={2n+1\choose 0}+{2n+1\choose 1}x+{2n+1\choose 2}x^2+\ldots\\ -(x-1)^{2n+1}&={2n+1\choose 0}x^{2n+1}-{2n+1\choose 1}x^{2n}+{2n+1\choose 2}x^{2n-1}-+\ldots\end{aligned}$$
Now the coefficient of $x^{2n+1}$ in $(1-x^2)^{2n+1}$ would be $0$, since only even degree terms are present in its expansion. So the answer should be $0$. Can anyone identify the error. Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{2n + 1}
\pars{-1}^{r}{2n + 1 \choose r}^{2}}
\\[5mm] = &\
\sum_{r = 0}^{2n + 1}
\pars{-1}^{\pars{2n + 1} - r}
\,\,\,{2n + 1 \choose \pars{2n + 1} - r}^{2}
\\[5mm] = &\
\color{red}{-}\sum_{r = 0}^{2n + 1}
\pars{-1}^{r}{2n + 1 \choose r}^{2}
\\[5mm] &\
\implies
\bbx{\sum_{r = 0}^{2n + 1}
\pars{-1}^{r}{2n + 1 \choose r}^{2} = 0} \\ &
\end{align}
