Let = _k=0^n((^nC_k)^2k+1) and = _k=0^n-1(^nC_k^nC_k+1k+2) If 5 =6 , then n equals _______…

Question

Let $\alpha =\sum _{k=0}^{n}(\frac{{({}^{n}{C}_{k})}^{2}}{k+1})$ and $\beta =\sum _{k=0}^{n-1}(\frac{{}^{n}{C}_{k}{}^{n}{C}_{k+1}}{k+2})$ If $5\alpha =6\beta $, then $n$ equals \_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

$\begin{gathered} \alpha = \sum _{k=0}^{n}\frac{{}^{n}{C}_{k}\cdot {}^{n}{C}_{k}}{k+1}\cdot \frac{n+1}{n+1} \ =\frac{1}{n+1}\sum _{k=0}^{n}{}^{n+1}{C}_{k+1}\cdot {}^{n}{C}_{n-k} \ \alpha =\frac{1}{n+1}\cdot {}^{2n+1}{C}_{n+1} \ \beta =\sum _{k=0}^{n-1}{C}_{k}\cdot \frac{{}^{n}{C}_{k+1}}{k+2}\frac{n+1}{n+1} \ \frac{1}{n+1}\sum _{k=0}^{n-1}{}^{n}{C}_{n-k}\cdot {}^{n+1}{C}_{k+2} \ =\frac{1}{n+1}\cdot {}^{2n+1}{C}_{n+2} \ \frac{\beta }{\alpha } =\frac{2n+1}{2n+1}{C}_{n+2} \ \frac{\beta }{\alpha } =\frac{2n+1-(n+2)+1}{n+2}=\frac{5}{6} \ n =10 \end{gathered}$

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