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माना $\alpha=\sum_{k=0}^{\mathrm{n}}\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ तथा $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$ हैं। यदि $5 \alpha=6 \beta$ हैं, तो $\mathrm{n}$ बराबर है ............
$6$
$7$
$9$
$10$
Solution
$\sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} $
$ =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} $
$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} $
$ =\sum_{k=0}^{n-1}{ }^n C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $
$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2} $
$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2}$
$\alpha=\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1}$
$ \beta=\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $
$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2}$
$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} $
$ \frac{\beta}{\alpha}=\frac{{ }^{2 n+1} C_{n+2}}{{ }^{2 n+1} C_{n+1}}=\frac{2 n+1-(n+2)+1}{n+2} $
$ \frac{\beta}{\alpha}=\frac{n}{n+2}=\frac{5}{6} $
$n=10$
