Cho n là số nguyên dương. \(\begin{equation} \frac{\mathrm{C}_{n}^{0}}{\mathrm{C}_{n+2}^{1}}+\frac{\mathrm{C}_{n}^{1}}{\mathrm{C}_{n+3}^{2}}+\frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n+4}^{3}}+\cdots+\frac{\mathrm{C}_{n}^{k}}{\mathrm{C}_{n+k+2}^{k+1}}+\cdots+\frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{2 n+2}^{n+1}} \end{equation}\) bằng với:
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Lời giải:
Báo sai\(\begin{equation} \frac{\mathrm{C}_{n}^{k}}{\mathrm{C}_{n+k+2}^{k+1}}=\frac{n !}{k !(n-k) !} \cdot \frac{(k+1) !(n+1) !}{(n+k+2) !}=\frac{n !(n+1) !(k+1)}{(n-k) !(n+k+2) !} \end{equation}\)
\(\begin{equation} =\frac{1}{2}\left[\frac{n !(n+1) !}{(n-k) !(n+k+1) !}-\frac{n !(n+1) !}{(n-k-1) !(n+k+2) !}\right] \end{equation}\)
\(\begin{equation} =\frac{1}{2}\left[\frac{\frac{(2 n+1) !}{n !(n+1) !}}{\frac{(2 n+1) !}{(n-k) !(n+k+1) !}}-\frac{\frac{(2 n+1) !}{n !(n+1) !}}{\frac{(2 n+1) !}{(n-k-1) !(n+k+2) !}}\right] \end{equation}\)
\(\begin{equation} =\frac{1}{2} \cdot \frac{\mathrm{C}_{2 n+1}^{n-k}-\mathrm{C}_{2 n+1}^{n-k-1}}{\mathrm{C}_{2 n+1}^{n}} \end{equation}\)
\(\begin{equation} \text { Với } k=0, \text { ta có } \frac{\mathrm{C}_{n}^{0}}{\mathrm{C}_{n+2}^{1}}=\frac{1}{2} \cdot \frac{\mathrm{C}_{2 n+1}^{n}-\mathrm{C}_{2 n+1}^{n-1}}{\mathrm{C}_{2 n+1}^{n}} \text { . } \end{equation}\)
\(\begin{equation} \text { Với } k=1, \text { ta có } \frac{C_{n}^{1}}{C_{n+3}^{2}}=\frac{1}{2} \cdot \frac{C_{2 n+1}^{n-1}-C_{2 n+1}^{n-2}}{C_{2 n+1}^{n}} \text { . } \end{equation}\)
...
\(\begin{equation} \text { Với } k=n-1, \text { ta có } \frac{\mathrm{C}_{n}^{n-1}}{\mathrm{C}_{2 n+1}^{n}}=\frac{1}{2} \cdot \frac{\mathrm{C}_{2 n+1}^{1}-\mathrm{C}_{2 n+1}^{0}}{\mathrm{C}_{2 n+1}^{n}} \text { . } \end{equation}\)
\(\begin{equation} \text { Với } k=n, \text { ta có } \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{2 n+2}^{n}}=\frac{1}{2} \cdot \frac{\mathrm{C}_{2 n+1}^{0}}{\mathrm{C}_{2 n+1}^{n}} \text { . } \end{equation}\)
Do đó: \(\begin{equation} \frac{\mathrm{C}_{n}^{0}}{\mathrm{C}_{n+2}^{1}}+\frac{\mathrm{C}_{n}^{1}}{\mathrm{C}_{n+3}^{2}}+\frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n+4}^{3}}+\cdots+\frac{\mathrm{C}_{n}^{k}}{\mathrm{C}_{n+k+2}^{k+1}}+\cdots+\frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{2 n+2}^{n+1}}=\frac{1}{2} \end{equation}\)