∑ k = 1 n ( − 1 ) k ( 2 n + 1 k ) = ∑ k = 1 n ( − 1 ) k ( 2 n k ) + ∑ k = 1 n ( − 1 ) k ( 2 n k − 1 ) ∑ k = 1 n ( − 1 ) k ( 2 n + 1 k ) = ( ( − 1 ) n ( 2 n n ) + ∑ k = 1 n − 1 ( − 1 ) k ( 2 n k ) ) + ∑ k = 0 n − 1 ( − 1 ) k + 1 ( 2 n k ) ∑ k = 1 n ( − 1 ) k ( 2 n + 1 k ) = − 1 + ( − 1 ) n ( 2 n n ) ∑ k = 0 n ( − 1 ) k ( 2 n + 1 k ) = ( − 1 ) n ( 2 n n ) ∑ k = 0 n ( − 1 ) k ( 2 n + 1 n − k ) = ( 2 n n ) _ _ ¯ ¯ {\displaystyle {\begin{aligned}&\sum \limits _{k=1}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n+1\\k\\\end{matrix}}\right)}=\sum \limits _{k=1}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n\\k\\\end{matrix}}\right)}+\sum \limits _{k=1}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n\\k-1\\\end{matrix}}\right)}\\&\sum \limits _{k=1}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n+1\\k\\\end{matrix}}\right)}=\left(\left(-1\right)^{n}\left({\begin{matrix}2n\\n\\\end{matrix}}\right)+\sum \limits _{k=1}^{n-1}{\left(-1\right)^{k}\left({\begin{matrix}2n\\k\\\end{matrix}}\right)}\right)+\sum \limits _{k=0}^{n-1}{\left(-1\right)^{k+1}\left({\begin{matrix}2n\\k\\\end{matrix}}\right)}\\&\sum \limits _{k=1}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n+1\\k\\\end{matrix}}\right)}=-1+\left(-1\right)^{n}\left({\begin{matrix}2n\\n\\\end{matrix}}\right)\\&\sum \limits _{k=0}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n+1\\k\\\end{matrix}}\right)=\left(-1\right)^{n}\left({\begin{matrix}2n\\n\\\end{matrix}}\right)}\\&{\overline {\overline {\underline {\underline {\sum \limits _{k=0}^{n}{\left(-1\right)^{k}\left({\begin{matrix}2n+1\\n-k\\\end{matrix}}\right)=\left({\begin{matrix}2n\\n\\\end{matrix}}\right)}}}}}}\\\end{aligned}}}
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