Let $\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1}$ and $\beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k}\binom{n}{k+1}}{k+2}$. If $5\alpha = 6\beta$, then $n$ equals $\underline{\hspace{1cm}}$.
Let $\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1}$ and $\beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k}\binom{n}{k+1}}{k+2}$. If $5\alpha = 6\beta$, then $n$ equals $\underline{\hspace{1cm}}$.