Let α = ∑ _ k = 0 ^ n ( (Ck n) ^ 2 k + 1 ) and β = ∑ _ k = 0 ^ n -...

Given,
$\alpha = \mathop{∑}_{k = 0}^{n}\left(\dfrac{{\left(Ck n\right)}^{2}}{k + 1}\right)$

$⇒ \alpha = \mathop{∑}_{k = 0}^{n}\dfrac{Ck n \cdot Ck n}{k + 1} \cdot \dfrac{n + 1}{n + 1}$

Now, using $Cr + 1 n + 1 = \dfrac{n + 1}{r + 1} \cdot Cr n$ we get,

$⇒ \alpha = \dfrac{1}{n + 1}\mathop{∑}_{k = 0}^{n} Ck + 1 n + 1 \cdot Cn - k n$

Now, using $\mathop{∑}_{k = 0}^{n} Cr + 1 n + 1 \cdot Cn - r n = Cn + 1 2n + 1$ we get,

$⇒ \alpha = \dfrac{1}{n + 1} \cdot Cn + 1 2n + 1 . . . . . . . . \left(i\right)$

Now, solving 

$\beta = \mathop{∑}_{k = 0}^{n - 1} Ck n \cdot \dfrac{Ck + 1 n}{k + 2}\dfrac{n + 1}{n + 1}$

$⇒ \beta = \dfrac{1}{n + 1}\mathop{∑}_{k = 0}^{n - 1} Cn - k n \cdot Ck + 2 n + 1$

Now, using $\mathop{∑}_{k = 0}^{n} Cr + 2 n + 1 \cdot Cn - r n = Cn + 2 2n + 1$

$⇒ \beta = \dfrac{1}{n + 1} \cdot Cn + 2 2n + 1 . . . . \left(ii\right)$

Now, dividing equation \left(i\right) \text{ and } \left(ii\right)$and using$\dfrac{Cr n}{Cr - 1 n} = \dfrac{n - r + 1}{r}$we get,</p><p>$\dfrac{\beta}{\alpha} = \dfrac{Cn + 2 2n + 1}{Cn + 1 2n + 1} = \dfrac{2n + 1 - ( n + 2 ) + 1}{n + 2}$</p><p>$⇒ \dfrac{\beta}{\alpha} = \dfrac{n}{n + 2} = \dfrac{5}{6}$</p><p>$⇒ n = 10