Question

The sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$ in the binomial expansion of $\left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9$ is:

• A. $\frac{21}{4}$
• B. $\frac{69}{16}$
• C. $\frac{63}{16}$
• D. $\frac{19}{4}$

Answer 1

Answer: (A)

$\frac{21}{4}$

Answer 2

Step 1. General Term of the Expansion:
The general term in the binomial expansion of $\left( x^{2/3} + \frac{1}{2}x^{-2/5} \right)^9$ is given by:

$T_{r+1} = \binom{9}{r} \left( x^{2/3} \right)^{9-r} \left( \frac{x^{-2/5}}{2} \right)^r$

Simplify the expression:

$T_{r+1} = \binom{9}{r} \left( \frac{1}{2} \right)^r x^{\left( \frac{6 - 2r}{3} - \frac{2r}{5} \right)}$

Step 2. For $x^{2/3}$:
Set the power of $x$ equal to $2/3$:

$\frac{6 - 2r}{3} - \frac{2r}{5} = \frac{2}{3}$

Solving this equation gives $r = 5$.
Substituting $r = 5$ into the coefficient formula:

$\text{Coefficient of } x^{2/3} = \binom{9}{5} \left( \frac{1}{2} \right)^5$

Step 3. For $x^{-2/5}$:
Set the power of $x$ equal to $-2/5$:

$\frac{6 - 2r}{3} - \frac{2r}{5} = -\frac{2}{5}$

Solving this equation gives $r = 6$.
Substituting $r = 6$ into the coefficient formula:

$\text{Coefficient of } x^{-2/5} = \binom{9}{6} \left( \frac{1}{2} \right)^6$

Step 4. Sum of the Coefficients:
Add the two coefficients:

$\text{Sum} = \binom{9}{5} \left( \frac{1}{2} \right)^5 + \binom{9}{6} \left( \frac{1}{2} \right)^6$

Simplify:$\text{Sum} = \frac{21}{4}$