JEE Main 2025 — 22 January, Evening Shift — Question 24

$\sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^{2}\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^{2}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}$

$=\sum_{r=1}^{30} r^{2}\left(\frac{31-r}{r}\right) \cdot \frac{30!}{r!(30-r)!}$

$\left(\because \frac{{ }^{30} \mathrm{C}_{\mathrm{r}}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\frac{30-\mathrm{r}+1}{\mathrm{r}}=\frac{31-\mathrm{r}}{\mathrm{r}}\right)$

$=\sum_{r=1}^{30} \frac{(31-r) 30!}{(r-1)!(30-r)!}$

$=30 \sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 29!}{(\mathrm{r}-1)!(30-\mathrm{r})!}$

$=30 \sum_{\mathrm{r}=1}^{30}(30-\mathrm{r}+1)^{29} \mathrm{C}_{30-\mathrm{r}}$

$=30\left(\sum_{\mathrm{r}=1}^{30}(31-\mathrm{r})^{29} \mathrm{C}_{30-\mathrm{r}}+\sum_{\mathrm{r}=1}^{30}{ }^{29} \mathrm{C}_{30-\mathrm{r}}\right)$

$=30\left(29 \times 2^{28}+2^{29}\right)=30(29+2) 2^{28}$

$=15 \times 31 \times 2^{29}$

$=465\left(2^{29}\right)$

$\alpha=465$