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Question

S=

$\frac{C^{0}_{n}}{C^{1}_{n+2}}$ +

$\frac{C^{1}_{n}}{C^{2}_{n+3}}$ +

$\frac{C^{2}_{n}}{C^{3}_{n+4}}$ +....+

$\frac{C^{k}_{n}}{C^{k+1}_{n+k+2}}$ +....+

$\frac{C^{n}_{n}}{C^{n+1}_{2n+2}}$ ( n

$\in$ N*)

score 2 · Answer 1

Ta có:

$\frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{n!(k+1)!(n+1)!}{k!(n-k)!(n+k+2)!}=\frac{n!(n+1)!}{(2n+1)!}.\frac{(k+1)(2n+1)!}{(n+k+2)!(n-k)!}=\frac{1}{C_{2n+1}^n}.\frac{(k+1)(2n+1)!}{(n+k+2)!(n-k)!}.$
Lại có:

$\frac{(k+1)(2n+1)!}{(n+k+2)!(n-k)!}=\frac{[(n+k+2)-(n-k)](2n+1)!}{2(n+k+2)!(n-k)!}=\frac{(2n+1)!}{2(n+k+1)!(n-k)!}-\frac{(2n+1)!}{2(n+k+2)!(n-k-1)!}=\frac{C_{2n+1}^{n+k+1}-C_{2n+1}^{n+k+2}}{2}$
Từ đó suy ra: