Chứng minh rằng
$\sum\limits_{k = 0}^n {\frac{C^{k}_{n}}{C^{k+1}_{n+k+2}} }=\frac{1}{2}  $

Bạn xem lại đề bài nhé. $C_k^n$ không tồn tại với $k –  Trần Nhật Tân 25-09-12 08:20 AM
Đặt $S(n)=\sum_{k=0}^{n}\frac{C_{n}^{k}}{C_{n+k+2}^{k+1}}$.
Trước tiên ta thấy
$\frac{C_{n}^{k}}{C_{n+k+2}^{k+1}}=\frac{\frac{n!}{(n-k)!k!}}{\frac{(n+k+2)!}{(k+1)!(n+1)!}}=\frac{n!(n+1)!(k+1)}{(n-k)!(n+k+2)!}=\frac{n!(n+1)!}{(2n+2)!} \times (k+1)C_{2n+2}^{n+k+2}$
Do đó
$S(n)= \frac{n!(n+1)!}{(2n+2)!} \sum_{k=0}^{n}\left( (k+1)C_{2n+2}^{n+k+2}\right) $
Mặt khác, đặt
$A=\sum_{k=0}^{n}\left((n+k+2)C_{2n+2}^{n+k+2}\right)$
$B=(n+1)\sum_{k=0}^{n}C_{2n+2}^{n+k+2}$
Thì
$A=\sum_{k=0}^{n}\left((2n+2)C_{2n+1}^{n+k+1}\right)=(2n+2) \times \frac{1}{2}2^{2n+1}$

$B=(n+1) \times \frac{1}{2}\left(2^{2n+2}-C_{2n+2}^{n+1}\right)$
Suy ra
$S(n)=\frac{n!(n+1)!}{(2n+2)!}(A-B)=\frac{n!(n+1)!}{(2n+2)!}\times \frac{1}{2}(n+1)C_{2n+2}^{n+1}=\frac{1}{2}$
đy qua đá vote cho bạn :) –  toiditimtoan 29-09-12 09:56 PM
Bài này không vote thì phí :) –  Lee 27-09-12 11:10 PM

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