Q28 / 75 · JEE Main 2025 (3 Apr Shift 1) · Mathematics
If $\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) .{ }^9 \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^9-\beta, \quad \alpha, \beta \in \mathrm{N}, \quad$ then $(\alpha+\beta)^2$ is equal to
- A$27$
- B$9$
- Correct answerC$81$
- D$18$
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 $(\alpha+\beta)^2=81$
Hence, the correct answer is option: 'C'.