Question

If \( a \) is the coefficient of \( x^7 \) and \( b \) is the term independent of \( x \) in the expansion of \( \left( \frac{3x^4}{4} - \frac{4}{3x^3} \right)^7 \), then \( \frac{b}{a} = \)

• A. \( -1 \)
• B. \( 1 \)
• C. \( -\frac{16}{9} \)
• D. \( \frac{16}{9} \)

Hint:

To simplify coefficients of the form \( (x/y)^k \cdot (y/x)^m \), combine them into \( (x/y)^{k-m} \) to avoid large intermediate calculations.

Answer 1

The Correct Option is C

Concept: The general term \( T_{r+1} \) in the expansion of \( (A + B)^n \) is given by \( ^nC_r A^{n-r} B^r \).
• \( T_{r+1} = ^7C_r \left( \frac{3x^4}{4} \right)^{7-r} \left( -\frac{4}{3x^3} \right)^r \)
• Separate constants and \( x \) terms to find specific coefficients.

Step 1: Simplifying the general term.
\[ T_{r+1} = ^7C_r \left( \frac{3}{4} \right)^{7-r} \left( -\frac{4}{3} \right)^r x^{4(7-r)} \cdot x^{-3r} = ^7C_r \left( \frac{3}{4} \right)^{7-r} \left( -\frac{4}{3} \right)^r x^{28-7r} \]

Step 2: Finding coefficient \( a \) (for \( x^7 \)).
Set \( 28 - 7r = 7 \implies 7r = 21 \implies r = 3 \). \[ a = ^7C_3 \left( \frac{3}{4} \right)^4 \left( -\frac{4}{3} \right)^3 = 35 \cdot \frac{3^4}{4^4} \cdot \left( -\frac{4^3}{3^3} \right) = 35 \cdot \frac{3}{4} \cdot (-1) = -\frac{105}{4} \]

Step 3: Finding coefficient \( b \) (independent of \( x \)).
Set \( 28 - 7r = 0 \implies 7r = 28 \implies r = 4 \). \[ b = ^7C_4 \left( \frac{3}{4} \right)^3 \left( -\frac{4}{3} \right)^4 = 35 \cdot \frac{3^3}{4^3} \cdot \frac{4^4}{3^4} = 35 \cdot \frac{4}{3} = \frac{140}{3} \]

Step 4: Calculating the ratio \( b/a \).
\[ \frac{b}{a} = \frac{140/3}{-105/4} = \frac{140}{3} \times \left( -\frac{4}{105} \right) = \frac{4 \times 35 \times 4}{-3 \times 3 \times 35} = -\frac{16}{9} \]