Sommes (E-M)

• Montrer que $\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ n-k\end{array}\right)=\left(\begin{array}{c}2n\\ n\end{array}\right)$.

Solution

$\begin{array}{cc}& \sum _{k=1}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ k\end{array}\right)=\sum _{k=1}^n{(-1)}^k\left(\begin{array}{c}2n\\ k\end{array}\right)+\sum _{k=1}^n{(-1)}^k\left(\begin{array}{c}2n\\ k-1\end{array}\right)\\ & \sum _{k=1}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ k\end{array}\right)=({(-1)}^n\left(\begin{array}{c}2n\\ n\end{array}\right)+\sum _{k=1}^{n-1}{(-1)}^k\left(\begin{array}{c}2n\\ k\end{array}\right))+\sum _{k=0}^{n-1}{(-1)}^{k+1}\left(\begin{array}{c}2n\\ k\end{array}\right)\\ & \sum _{k=1}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ k\end{array}\right)=-1+{(-1)}^n\left(\begin{array}{c}2n\\ n\end{array}\right)\\ & \sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ k\end{array}\right)={(-1)}^n\left(\begin{array}{c}2n\\ n\end{array}\right)\\ & \stackrel{‾}{\stackrel{‾}{\underset{\_}{\underset{\_}{\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}2n+1\\ n-k\end{array}\right)=\left(\begin{array}{c}2n\\ n\end{array}\right)}}}}\\ \end{array}$