Combination 03

Chứng minh rằng với mọi $n\in N$, ta luôn có:

$C_{2n+2}^{n+2}+2C_{2n+2}^{n+3}+3C_{2n+2}^{n+4}+...+\left(n+1\right)C_{2n+2}^{2n+2}$$=\frac{1}{2}\left(n+2\right)C_{2n+2}^n$

Lời giải

$\frac{C_n^k}{C_{n+2+k}^{k+1}}$$=\frac{(n+1)(C_{2n+1}^{n-k}-C_{2n+1}^{n-k-1})}{(n+2)C_{2n+2}^n}$$=\frac{(k+1)C_{2n+2}^{n+k+2}}{(n+2)C_{2n+2}^n}$

Khi đó 

$$\begin{array}{rl}& \frac{1}{(n+2)C_{2n+2}^n}\sum _{k=0}^n(k+1)C_{2n+2}^{n+k+2}=\sum _{k=0}^n\frac{C_n^k}{C_{n+2+k}^{k+1}}\\ \iff & \frac{1}{(n+2)C_{2n+2}^n}\sum _{k=0}^n(k+1)C_{2n+2}^{n+k+2}=\frac{1}{2}\\ \iff & \sum _{k=0}^n(k+1)C_{2n+2}^{n+k+2}=\frac{1}{2}(n+2)C_{2n+2}^n\end{array}$$