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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:
$\frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{1}{2C_{2n+1}^n}(C_{2n+1}^{n+k+1}-C_{2n+1}^{n+k+2})$.
Lấy tổng theo $k$ từ $0$ đến $n$ ta được:
$S=\frac{1}{2C_{2n+1}^n}(C_{2n+1}^{n+1}-C_{2n+1}^{2n+2})=\frac{1}{2}.$
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