Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& \sum _{\mathrm{r}=1}^{30}\frac{{\mathrm{r}}^2{\left({}^{30}{\mathrm{C}}_{\mathrm{r}}\right)}^2}{{}^{30}{\mathrm{C}}_{\mathrm{r}-1}}\\ & =\sum _{\mathrm{r}=1}^{30}{\mathrm{r}}^2\left(\frac{31-\mathrm{r}}{\mathrm{r}}\right)\cdot \frac{30!}{\mathrm{r}!(30-\mathrm{r})!}\\ & \left(\because \frac{{}^{30}{\mathrm{C}}_{\mathrm{r}}}{{}^{30}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{30-\mathrm{r}+1}{\mathrm{r}}=\frac{31-\mathrm{r}}{\mathrm{r}}\right)\end{array}$ $\begin{array}{rl}& =\sum _{\mathrm{r}=1}^{30}\frac{(31-\mathrm{r})30!}{(\mathrm{r}-1)!(30-\mathrm{r})!}\\ & =30\sum _{\mathrm{r}=1}^{30}\frac{(31-\mathrm{r})29!}{(\mathrm{r}-1)!(30-\mathrm{r})!}\\ & =30\sum _{\mathrm{r}=1}^{30}(30-\mathrm{r}+1{)}^{29}{\mathrm{C}}_{30-\mathrm{r}}\\ & =30\left(\sum _{\mathrm{r}=1}^{30}(31-\mathrm{r}{)}^{29}{\mathrm{C}}_{30-\mathrm{r}}+\sum _{\mathrm{r}=1}^{30}{}^{29}{\mathrm{C}}_{30-\mathrm{r}}\right)\\ & =30\left(29\times 2^{28}+2^{29}\right)=30(29+2)2^{28}\\ & =15\times 31\times 2^{29}\\ & =465\left(2^{29}\right)\\ & \alpha =465\end{array}$