JEE Main 2025 — 3 April, Morning Shift — Question 28

Given that $\sum_{\mathrm{r}=1}^{9}\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^{9} \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^{9}-\beta, \alpha, \beta \in \mathrm{N}$

Now, $\sum_{\mathrm{r}=1}^{9}\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^{9} \mathrm{C}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^{9}\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right) \cdot{ }^{9} \mathrm{C}_{\mathrm{r}}+\sum_{\mathrm{r}=1}^{9}\left(\frac{3}{2^{\mathrm{r}}}\right) \cdot{ }^{9} \mathrm{C}_{\mathrm{r}}$

$=\sum_{\mathrm{r}=1}^{9}\left(\frac{9}{2^{\mathrm{r}}}\right) \cdot{ }^{8} \mathrm{C}_{\mathrm{r}-1}+3 \sum_{\mathrm{r}=1}^{9}{ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\left[\mathrm{U} \sin \mathrm{g} \frac{{ }^{9} \mathrm{C}_{\mathrm{r}}}{{ }^{8} \mathrm{C}_{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right]$

$=\frac{9}{2} \sum_{\mathrm{r}=1}^{9}{ }^{8} \mathrm{C}_{\mathrm{r}-1}\left(\frac{1}{2}\right)^{\mathrm{r}-1}+3\left(\sum_{\mathrm{r}=0}^{9}\left({ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\right)-1\right)$

$=\frac{9}{2}\left(1+\frac{1}{2}\right)^{8}+3\left(\left(1+\frac{1}{2}\right)^{9}-1\right)$

$=\frac{9}{2} \cdot\left(\frac{3}{2}\right)^{8}+3\left(\frac{3}{2}\right)^{9}-3=6 \cdot\left(\frac{3}{2}\right)^{9}-3$

Hence, $\alpha=6, \beta=3$

Thus $(\alpha+\beta)^{2}=81$.