r=1∑9(2rr+3)⋅9Cr=α(23)9−β,α,β∈N Now, $\begin{aligned}
& \sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}{\mathrm{r}}=\sum{\mathrm{r}=1}^9\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}{\mathrm{r}}+\sum{\mathrm{r}=1}^9\left(\frac{3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}{\mathrm{r}} \
& =\sum{\mathrm{r}=1}^9\left(\frac{9}{2^{\mathrm{r}}}\right) \cdot{ }^8 \mathrm{C}{\mathrm{r}-1}+3 \sum{\mathrm{r}=1}^9{ }^9 \mathrm{C}{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\left[\mathrm{U} \sin \mathrm{~g} \frac{{ }^9 \mathrm{C}{\mathrm{r}}}{{ }^8 \mathrm{C}{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right] \
& =\frac{9}{2} \sum{\mathrm{r}=1}^9{ }^8 \mathrm{C}{\mathrm{r}-1}\left(\frac{1}{2}\right)^{\mathrm{r}-1}+3\left(\sum{\mathrm{r}=0}^9\left({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\right)-1\right) \
& =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left(\left(1+\frac{1}{2}\right)^9-1\right)
\end{aligned}=\frac{9}{2} \cdot\left(\frac{3}{2}\right)^8+3\left(\frac{3}{2}\right)^9-3=6 \cdot\left(\frac{3}{2}\right)^9-3Hence,\alpha=6, \beta=3Thus(\alpha+\beta)^2=81$
If Σ_ =1^9 ( +3 2^ ) . ^9 C_ =α ( 3 2 )^9-β, α, β ∈ N, then (α+β)^2 is equal to
JEE Main 2025 — Mathematics Algebra
If r=1∑9(2rr+3).9Cr=α(23)9−β,α,β∈N, then (α+β)2 is equal to
Held on 3 Apr 2025 · Verified 6 Jul 2026.
Options
- A
27
- B
9
- C
81
- D
18
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