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 ______

Answer 1

Given :
Coefficient of x⁷ is \((ax-\frac{1}{bx})^{13}\)
\(T_{r+1}={^{13}C_r}(ax)^{13-r}(-\frac{1}{bx^2})^r\)
13 - 3r = 7
 ⇒ r = 2
Coefficient = \({^{13}C_2}\frac{a^{11}}{b^2}\)
Coefficient of x^-5 is \((ax+\frac{1}{bx^2})^{13}\)
\(T_{r+1}={^{13}C_r}(ax)^{13-r}(\frac{1}{bx^2})^r\)
13 - 3r = -5
⇒ r = 6
Coefficient = \({^{13}C_6}\frac{a^7}{b^6}\)
Now,
\({^{13}C_2}\frac{a^{11}}{b^2}={^{13}C_6\frac{a^7}{b^6}}\)
\(a^4b^4=\frac{^{13}C_6}{^{13}C_2}=22\)
So, the correct answer is 22.