JEE Main 2025 — Binomial Theorem Question with Solution

${\sum }_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right)\cdot {}^9{\mathrm{C}}_{\mathrm{r}}=\alpha {\left(\frac{3}{2}\right)}^9-\beta ,\alpha ,\beta \in \mathrm{N}$ Now, $\begin{aligned} & \sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}+\sum_{\mathrm{r}=1}^9\left(\frac{3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}} \\ & =\sum_{\mathrm{r}=1}^9\left(\frac{9}{2^{\mathrm{r}}}\right) \cdot{ }^8 \mathrm{C}_{\mathrm{r}-1}+3 \sum_{\mathrm{r}=1}^9{ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\left[\mathrm{U} \sin \mathrm{~g} \frac{{ }^9 \mathrm{C}_{\mathrm{r}}}{{ }^8 \mathrm{C}_{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right] \\ & =\frac{9}{2} \sum_{\mathrm{r}=1}^9{ }^8 \mathrm{C}_{\mathrm{r}-1}\left(\frac{1}{2}\right)^{\mathrm{r}-1}+3\left(\sum_{\mathrm{r}=0}^9\left({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\right)-1\right) \\ & =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left(\left(1+\frac{1}{2}\right)^9-1\right) \end{aligned}$ $=\frac{9}{2}\cdot {\left(\frac{3}{2}\right)}^8+3{\left(\frac{3}{2}\right)}^9-3=6\cdot {\left(\frac{3}{2}\right)}^9-3$ Hence, $\alpha =6,\beta =3$ Thus $(\alpha +\beta {)}^2=81$