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