Binomial Theorem Question - None

Question

If the coefficient of x⁷ in $\left(a x-\frac{1}{b x^2}\right)^{13}$ and the coefficient of x^-5 in $\left(a x+\frac{1}{b x^2}\right)^{13}$ are equal, then a⁴b⁴ is equal to:

Accepted Answer

The general term is T_r+1 = ¹³C_r(ax)^13-r$\left(-\frac{1}{b x^2}\right)^r$
= ¹³C_r(a)^13-r$\left(-\frac{1}{b}\right)^r$x^13-3r
Since 13 - 3r = 7 $\Rightarrow$ r = 2
Coefficient of x⁷ = ¹³C₂(a)¹¹$\cdot$$\frac{1}{b^2}$
Now, T_r+1 = ¹³C_r(ax)^13-r$\left(\frac{1}{b x^2}\right)^{\mathrm{r}}$
13 - 3r = -5 $\Rightarrow$ r = 6
Coefficient of x^-5 = ¹³C₆(a)⁷$\cdot$$\frac{1}{b^6}$
Since, ${ }^{13} C_2 \frac{a^{11}}{b^2}={ }^{13} C_6 \frac{a^7}{b^6}$ $\Rightarrow$ a⁴b⁴ = $\frac{{ }^{13} C_6}{{ }^{13} C_2}$ = 22

Question:

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