Let alpha=sum_{k=0}^nleft(frac{left({ }^n C_k | vedclass

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Question

$\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} $

$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

$\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} $

$n=10$

Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals

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