Question

The coefficient of \( x^3 \) in the expansion of \( (1 - 2x)^{\frac{1}{2}} (1 + 3x)^{\frac{1}{3}} \) is:

• A. \( \mathbf{\frac{-20}{3}} \)
• B. \( \frac{20}{3} \)
• C. \( \frac{17}{3} \)
• D. \( \frac{-17}{3} \)

Hint:

To find a specific coefficient in a product of binomials, combine terms from both expansions whose exponents sum to the desired power.

Answer 1

The Correct Option is A

We expand both expressions using the Binomial Theorem. We want the coefficient of \( x^3 \) in the product. Let: \[ (1 - 2x)^{1/2} = \sum_{n=0}^{\infty} \binom{1/2}{n} (-2x)^n, \quad (1 + 3x)^{1/3} = \sum_{m=0}^{\infty} \binom{1/3}{m} (3x)^m \] We want combinations where \( n + m = 3 \). Valid (n, m) pairs: (0,3), (1,2), (2,1), (3,0) Calculate: \[ \text{Term 1: } \binom{1/2}{0}(-2x)^0 \cdot \binom{1/3}{3}(3x)^3 = 1 \cdot \frac{1/3 \cdot (-2/3) \cdot (-5/3)}{6} \cdot 27x^3 \] \[ \text{Term 2: } \binom{1/2}{1}(-2x)^1 \cdot \binom{1/3}{2}(3x)^2 = \frac{1}{2}(-2x) \cdot \frac{1/3 \cdot (-2/3)}{2} \cdot 9x^2 \] \[ \text{Term 3: } \binom{1/2}{2}(-2x)^2 \cdot \binom{1/3}{1}(3x)^1 = \frac{(1/2)(-1/2)}{2} \cdot 4x^2 \cdot \frac{1}{3} \cdot 3x \] \[ \text{Term 4: } \binom{1/2}{3}(-2x)^3 \cdot \binom{1/3}{0}(3x)^0 \] Adding these gives final coefficient \( \boxed{\frac{-20}{3}} \).