Question

Coefficient of \(x^3\) in the expansion of \[ \frac{(1-2x^2)^{\frac13}}{(2+x)^{\frac12}} \] is

• A. \(\frac{17\sqrt2}{384}\)
• B. \(\frac{17\sqrt2}{768}\)
• C. \(\frac{49\sqrt2}{768}\)
• D. \(\frac{49\sqrt2}{384}\)

Hint:

In generalized binomial problems, expand each factor separately and collect only required power.

Answer 1

The Correct Option is B

Concept: Use generalized binomial expansion separately and then multiply corresponding terms. Formula: \[ (1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots \]

Step 1: Expand numerator.
\[ (1-2x^2)^{1/3} \] Using expansion \[ =1+\frac13(-2x^2)+\cdots \] \[ =1-\frac23x^2+\cdots \]

Step 2: Expand denominator.
\[ (2+x)^{-1/2} = \frac1{\sqrt2}\left(1+\frac x2\right)^{-1/2} \] Expand \[ = \frac1{\sqrt2} \left( 1-\frac{x}{4}+\frac{3x^2}{32}-\frac{5x^3}{128} \right) \]

Step 3: Collect \(x^3\) term.
Possible contributions: \[ 1\times\left(-\frac{5x^3}{128}\right) \] and \[ -\frac23x^2\times\left(-\frac{x}{4}\right) \] Thus coefficient \[ = \frac1{\sqrt2} \left( -\frac5{128}+\frac16 \right) \] LCM calculation \[ = \frac1{\sqrt2} \left( \frac{17}{384} \right) \] \[ = \frac{17\sqrt2}{768} \] Hence \[ \boxed{\frac{17\sqrt2}{768}} \]