If $\sum\limits_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathbb{N}$, then $(\alpha+\beta)^2$ is equal to

<p>$$\begin{aligned} & \sum_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\sum_{r=1}^9 \frac{r}{2^r} \cdot \frac{9}{r} \cdot{ }^8 C_{r-1}+\sum_{r=1}^9 3 \cdot{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=1}^9{ }^8 C_{r-1}\left(\frac{1}{2}\right)^{r-1}+3 \sum_{r=1}^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=0}^8{ }^8 C_{r-1}\left(\frac{1}{2}\right)^r+3 \sum_{r=1}^9{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left[\left(1+\frac{1}{2}\right)^9-{ }^9 C_0\left(\frac{1}{2}\right)^0\right] \\ & =\frac{9}{2} \cdot \frac{3^8}{2^8}+3\left[\frac{3^9}{2^9}-1\right] \\ & =\frac{3^{10}}{2^9}+\frac{3^{10}}{2^9}-3=4 \cdot \frac{3^{10}}{2^{10}}-3 \\ & =4\left(\frac{3}{2}\right)^{10}-3 \\ & =6\left(\frac{3}{2}\right)^9-3 \\ & \alpha=6, \beta=3 \Rightarrow(\alpha+\beta)^2=81 \end{aligned}$$</p>