If $\sum_\limits{r=1}^{30} \frac{r^2\left({ }^{30} C_r\right)^2}{{ }^{30} C_{r-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to _________.

<p>$$\begin{aligned} & \sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^2\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^2}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}} \\ & =\sum_{\mathrm{r}=1}^{30} \mathrm{r}^2\left(\frac{31-\mathrm{r}}{\mathrm{r}}\right) \cdot \frac{30!}{\mathrm{r}!(30-\mathrm{r})!} \\ & \left(\because \frac{{ }^{30} \mathrm{C}_{\mathrm{r}}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\frac{30-\mathrm{r}+1}{\mathrm{r}}=\frac{31-\mathrm{r}}{\mathrm{r}}\right) \\ & =\sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 30!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 29!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30}(30-\mathrm{r}+1)^{29} \mathrm{C}_{30-\mathrm{r}} \\ & =30\left(\sum_{\mathrm{r}=1}^{30}(31-\mathrm{r})^{29} \mathrm{C}_{30-\mathrm{r}}+\sum_{\mathrm{r}=1}^{30}{ }^{29} \mathrm{C}_{30-\mathrm{r}}\right) \\ & =30\left(29 \times 2^{28}+2^{29}\right)=30(29+2) 2^{28} \\ & =15 \times 31 \times 2^{29} \\ & =465\left(2^{29}\right) \\ & \alpha=465 \end{aligned}$$</p>