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.