Question

The coefficient of $ x^2 $ in the expansion of \[ (1 - 3x)^{\frac{-1}{4}} \] is

• A. \( \frac{45}{64} \)
• B. \( \frac{45}{8} \)
• C. \( \frac{45}{16} \)
• D. \( \frac{45}{32} \)

Hint:

In the binomial expansion for fractional exponents, carefully apply the general term formula and calculate the binomial coefficients to find the desired power.

Answer 1

The Correct Option is D

Step 1: Use the binomial expansion formula for the expansion of \( (1 - 3x)^{\frac{-1}{4}} \). The binomial expansion of \( (1 + u)^n \) is given by: \[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots \] where \( u = -3x \) and \( n = \frac{-1}{4} \).
Step 2: We need to find the coefficient of \( x^2 \). The general term of the expansion is: \[ T_k = \binom{n}{k} u^k \] For the \( x^2 \) term, we substitute \( k = 2 \), \( u = -3x \), and \( n = \frac{-1}{4} \). \[ T_2 = \binom{\frac{-1}{4}}{2} (-3x)^2 = \binom{\frac{-1}{4}}{2} 9x^2 \] Step 3: The binomial coefficient \( \binom{\frac{-1}{4}}{2} \) is calculated as: \[ \binom{\frac{-1}{4}}{2} = \frac{\left( \frac{-1}{4} \right) \left( \frac{-5}{4} \right)}{2!} = \frac{\frac{5}{16}}{2} = \frac{5}{32} \] Step 4: Therefore, the coefficient of \( x^2 \) is: \[ \frac{5}{32} \times 9 = \frac{45}{32} \]