If , then is equal to [2025]

Question

If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

Smart Achievers · Accepted Answer

(3)

We have, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$

Now, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{r}{2^{r}}) \cdot {}^{9}C_{r} + \sum_{r = 1}^{9} (\frac{3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{9}{2^{r}}) \cdot {}^{8}C_{r - 1} + 3 \sum_{r = 1}^{9} {}^{9}C_{r} {(\frac{1}{2})}^{r}$

$= \frac{9}{2} \sum_{r = 1}^{9} {}^{8}C_{r - 1} {(\frac{1}{2})}^{r - 1} + 3 (\sum_{r = 0}^{9} ({}^{9}C_{r} {(\frac{1}{2})}^{r}) - 1)$

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

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

$= 6 \cdot {(\frac{3}{2})}^{9} - 3$

On comparing, we get

$\Rightarrow α = 6 and β = 3$

$∴ {(α + β)}^{2} = 81$ .

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