Let α = ∑ k = 0 n C k n 2 k + 1 and β = ∑ k = 0 n - 1 C k n C k + 1 n k + 2 . If 5 α = 6 β , then n equals

Given,
$\alpha =\sum _{k=0}^n\left(\frac{{\left({}^n\mathrm{C}_k\right)}^2}{k+1}\right)$

$\Rightarrow \alpha =\sum _{k=0}^n\frac{{}^{n}C_k·{}^{n}C_k}{k+1}·\frac{n+1}{n+1}$

Now, using ${}^{n+1}C_{r+1}=\frac{n+1}{r+1}·{}^{n}C_r$ we get,

$\Rightarrow \alpha =\frac{1}{n+1}\sum _{k=0}^n{}^{n+1}C_{k+1}·{}^{n}C_{n-k}$

Now, using $\sum _{k=0}^n{}^{n+1}C_{r+1}·{}^{n}C_{n-r}={}^{2n+1}C_{n+1}$ we get,

$\Rightarrow \alpha =\frac{1}{n+1}·{}^{2n+1}C_{n+1} ........\left(i\right)$

Now, solving 

$\beta =\sum _{k=0}^{n-1}{}^{n}C_k·\frac{{}^{n}C_{k+1}}{k+2}\frac{n+1}{n+1}$

$\Rightarrow \beta =\frac{1}{n+1}\sum _{k=0}^{n-1}{}^{n}C_{n-k}·{}^{n+1}C_{k+2}$

Now, using $\sum _{k=0}^n{}^{n+1}C_{r+2}·{}^{n}C_{n-r}={}^{2n+1}C_{n+2}$

$\Rightarrow \beta =\frac{1}{n+1}·{}^{2n+1}C_{n+2} ....\left(ii\right)$

Now, dividing equation $\left(i\right) \& \left(ii\right)$ and using $\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}$ we get,

$\frac{\beta }{\alpha }=\frac{{}^{2n+1}C_{n+2}}{{}^{2n+1}C_{n+1}}=\frac{2n+1-(n+2)+1}{n+2}$

$\Rightarrow \frac{\beta }{\alpha }=\frac{n}{n+2}=\frac{5}{6}$

$\Rightarrow n=10$