Question

If the coefficient of x⁷ in expansion of \((ax- \frac{1}{bx^2})^{13}\) is equal to the coefficient of x^-5 in expansion of \((ax + \frac{1}{bx^2})13\), then a⁴b⁴ is ______

• A. 22
• B. 44
• C. 11
• D. 33

Answer 1

The Correct Option is A

Given:

The general term of the binomial expansion is:

\( T_{r+1} = \binom{13}{r} a^{13-r} \left(-\frac{1}{b}\right)^r x^{13 - 3r} \) \quad \((1)\)

• Step 1: Coefficient of \( x^7 \):

  \( 13 - 3r = 7 \implies r = 2 \).

  Substitute \( r = 2 \) into (1):

  \( T_3 = \binom{13}{2} a^{11} \left(-\frac{1}{b}\right)^2 x^7 \).

  Simplify:

  Coefficient of \( x^7 = \binom{13}{2} \frac{a^{11}}{b^2} \).

• Step 2: Coefficient of \( x^{-5} \):

  \( 13 - 3r = -5 \implies r = 6 \).

  Substitute \( r = 6 \) into (1):

  \( T_7 = \binom{13}{6} a^7 \left(-\frac{1}{b}\right)^6 x^{-5} \).

  Simplify:

  Coefficient of \( x^{-5} = \binom{13}{6} \frac{a^7}{b^6} \).

• Step 3: Condition: Coefficient of \( x^7 = x^{-5} \):

  \( \binom{13}{2} \frac{a^{11}}{b^2} = \binom{13}{6} \frac{a^7}{b^6} \).

  Simplify:

  \( \frac{a^{11}}{b^2} = \frac{\binom{13}{6}}{\binom{13}{2}} \cdot \frac{a^7}{b^6} \implies a^4 \cdot b^4 = \frac{\binom{13}{6}}{\binom{13}{2}} \).

• Step 4: Simplify binomial coefficients:

  \( \binom{13}{6} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{720} \).

  \( \binom{13}{2} = \frac{13 \cdot 12}{2} = 78 \).

  Substitute:

  \( a^4 \cdot b^4 = \frac{\binom{13}{6}}{\binom{13}{2}} = \frac{\frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{720}}{78} = 22 \).

Final Answer: \( a^4 \cdot b^4 = 22 \).