Question

If $ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta $, $ \alpha, \beta \in \mathbb{N} $, then $ (\alpha + \beta)^2 $ is equal to

• A. 27
• B. 9
• C. 81
• D. 18

Hint:

Use the identity \( r \cdot {^nC_r} = n \cdot {^{n-1}C_{r-1}} \) to simplify the summation. Also, remember the binomial expansion \( (1+x)^n = \sum_{r=0}^{n} {^nC_r} x^r \).

Answer 1

The Correct Option is C

Given:
\[ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta,\quad \alpha, \beta \in \mathbb{N} \] We split the sum as: \[ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \sum_{r=1}^{9} \left( \frac{r}{2^r} \cdot {^9C_r} \right) + \sum_{r=1}^{9} \left( \frac{3}{2^r} \cdot {^9C_r} \right) \] Now, use the identity: \[ \frac{r}{2^r} \cdot {^9C_r} = \frac{9}{2^r} \cdot {^8C_{r-1}} \] So, \[ \sum_{r=1}^{9} \frac{r}{2^r} \cdot {^9C_r} = 9 \sum_{r=1}^{9} \frac{1}{2^r} \cdot {^8C_{r-1}} = \frac{9}{2} \sum_{r=1}^{9} {^8C_{r-1}} \cdot \left( \frac{1}{2} \right)^{r-1} \] Make the substitution \( s = r-1 \Rightarrow s = 0 \) to \( 8 \): \[ = \frac{9}{2} \sum_{s=0}^{8} {^8C_s} \left( \frac{1}{2} \right)^s = \frac{9}{2} \cdot \left(1 + \frac{1}{2}\right)^8 = \frac{9}{2} \cdot \left( \frac{3}{2} \right)^8 \] Now for the second term: \[ \sum_{r=1}^{9} \frac{3}{2^r} \cdot {^9C_r} = 3 \left( \sum_{r=0}^{9} {^9C_r} \cdot \left( \frac{1}{2} \right)^r - {^9C_0} \cdot 1 \right) = 3 \left( \left( 1 + \frac{1}{2} \right)^9 - 1 \right) = 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right) \] Adding both parts: \[ \frac{9}{2} \cdot \left( \frac{3}{2} \right)^8 + 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right) = \left( \frac{9}{2} \cdot \frac{2}{3} + 3 \right) \cdot \left( \frac{3}{2} \right)^9 - 3 = (3 + 3) \cdot \left( \frac{3}{2} \right)^9 - 3 = 6 \left( \frac{3}{2} \right)^9 - 3 \]
Thus, \( \alpha = 6 \), \( \beta = 3 \) \[ \therefore (\alpha + \beta)^2 = (6 + 3)^2 = 81 \]