Let $\alpha=\sum_\limits{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\beta=\left(\sum_\limits{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $140<\frac{2 \alpha}{\beta}<281$, then the value of $n$ is _________.

<p>$$\begin{aligned} \alpha= & \sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r \\ & =4 \sum_{r=0}^n r^2{ }^n C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\ & =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\ & =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\ & \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1}{n+1}\right) \\ & (1+x)^n=\sum_{r=0}^n{ }^n C_r x^r \end{aligned}$$</p> <p>$$\begin{aligned} & \int_\limits0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_\limits{r=0} \frac{{ }^n C}{r+1} \\ & \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\ \Rightarrow & \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\ \Rightarrow & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\ \Rightarrow & (n+1)^3=216 \\ \Rightarrow & n+1=6 \Rightarrow n=5 \end{aligned}$$</p>