Question

If the sum, \( \sum_{r=1}^9 \frac{r+3}{2r} \cdot \binom{9}{r} \) is equal to \( \alpha \cdot \left( \frac{3}{2} \right)^9 - \beta \), then the value of \( (\alpha + \beta)^2 \) is equal to:

• A. 81
• B. 9
• C. 36
• D. 27

Hint:

Binomial sums often break into simpler sums, using known binomial identities to simplify.

Answer 1

The Correct Option is B

The sum is represented as: \[ \sum_{r=1}^9 \frac{r+3}{2r} \cdot \binom{9}{r} \] Separating the terms yields: \[ \sum_{r=1}^9 \left( \frac{r}{2r} + \frac{3}{2r} \right) \cdot \binom{9}{r} \] This simplifies to: \[ \sum_{r=1}^9 \frac{1}{2} \binom{9}{r} + \frac{3}{2} \sum_{r=1}^9 \frac{1}{r} \binom{9}{r} \] Applying binomial sum properties and simplifying, we determine that \( (\alpha + \beta)^2 = 9 \).