If $\sum_{r=1}^{9}\left(\frac{r+3}{2^{r}}\right) {}^{9}C_{r} = \alpha\left(\frac{3}{2}\right)^{9}-\beta$, where $\alpha,\beta\in\mathbb{N}$, then $(\alpha+\beta)^{2}$ is equal to
- (1) 27
- (2) 9
- (3) 81
- (4) 18
Solution:
$$ \sum_{r=1}^{9} \left( \frac{r+3}{2^{r}} \right) {}^{9}C_{r} $$
$$ = \sum_{r=1}^{9} \frac{r}{2^{r}} {}^{9}C_{r} + \sum_{r=1}^{9} \frac{3}{2^{r}} {}^{9}C_{r} $$
$$ {}^{n}C_{r} = \frac{n}{r} {}^{n-1}C_{r-1} $$
$$ = \sum_{r=1}^{9} \frac{r}{2^{r}} \left( \frac{9}{r} {}^{8}C_{r-1} \right) + 3 \sum_{r=1}^{9} {}^{9}C_{r} \left(\frac{1}{2}\right)^{r} $$
$$ = 9 \sum_{r=1}^{9} {}^{8}C_{r-1} \frac{1}{2^{r}} + 3 \left( \sum_{r=0}^{9} {}^{9}C_{r} \left(\frac{1}{2}\right)^{r} – {}^{9}C_{0} \left(\frac{1}{2}\right)^{0} \right) $$
$$ = \frac{9}{2} \sum_{r=1}^{9} {}^{8}C_{r-1} \left(\frac{1}{2}\right)^{r-1} + 3 \left( \sum_{r=0}^{9} {}^{9}C_{r} \left(\frac{1}{2}\right)^{r} – 1 \right) $$
$$ (1+x)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} x^{r} $$
$$ = \frac{9}{2} \left( 1 + \frac{1}{2} \right)^{8} + 3 \left( \left(1 + \frac{1}{2}\right)^{9} – 1 \right) $$
$$ = \frac{9}{2} \left( \frac{3}{2} \right)^{8} + 3 \left( \frac{3}{2} \right)^{9} – 3 $$
$$ = 3 \cdot \frac{3}{2} \cdot \left( \frac{3}{2} \right)^{8} + 3 \left( \frac{3}{2} \right)^{9} – 3 $$
$$ = 3 \left( \frac{3}{2} \right)^{9} + 3 \left( \frac{3}{2} \right)^{9} – 3 $$
$$ = 6 \left( \frac{3}{2} \right)^{9} – 3 $$
$$ \alpha\left(\frac{3}{2}\right)^{9} – \beta = 6 \left( \frac{3}{2} \right)^{9} – 3 $$
$$ \alpha = 6, \quad \beta = 3 $$
$$ (\alpha+\beta)^{2} = (6+3)^{2} = 9^{2} = 81 $$
Ans. (3)