Official solution
\begin{gathered} \alpha = \sum _{r=0}^{n}(4{r}^{2}+2r+1){}^{n}{C}_{r} \\ =4\sum _{r=0}^{n}{r}^{2}{}^{n}{C}_{r}+2\sum _{r=0}^{n}r\cdot {}^{n}{C}_{r}+\sum _{r=0}^{n}{}^{n}{C}_{r} \\ =4n(n+1){2}^{n-2}+2\cdot n\cdot {2}^{n-1}+{2}^{n} \\ ={2}^{n}(n(n+1)+n+1)={2}^{n}(n+1{)}^{2} \\ \beta =\sum _{r=0}^{n}(\frac{{}^{n}{C}_{r}}{r+1})+(\frac{1}{n+1}) \\ (1+x{)}^{n}=\sum _{r=0}^{n}{}^{n}{C}_{r}{x}^{r}
\int _{0}^{1}(1+x{)}^{n}dx={\sum _{r=0}^{n}\frac{{}^{n}{C}_{r}{x}^{r+1}}{r+1}|}_{1}^{0}=\sum _{r=0}\frac{{}^{n}C}{r+1} \\ {\frac{(1+x{)}^{+1}}{n+1}|}_{1}^{0}=\frac{{2}^{n}-1}{n+1} \\ \Rightarrow \beta =\frac{{2}^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\ \Rightarrow \frac{2\alpha }{\beta }=\frac{{2}^{n+1}(n\, 1)}{(\frac{{2}^{n+1}}{n+1})}=(n+1{)}^{3}\in (140,281) \\ \Rightarrow (n+1{)}^{3}=216 \\ \Rightarrow n+1=6\Rightarrow n=5 \end{gathered}