If the coefficient of x ^7 in left(a x-frac{1 | vedclass

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Question

$T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r$

$={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}$

$1 3 - 3 r = 7

Vedclass · Accepted Answer

$22$

Vedclass · Answer

$T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r$

$={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}$

$1 3 - 3 r = 7 \Rightarrow r=2$

Coefficient of $x^7={ }^{13} C_2(a)^{11} \cdot \frac{1}{b^2}$

In the other expansion $T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r$

$13-3 r=-5 \Rightarrow r=6$

Coefficient of $x^{-5}={ }^{13} C_6(a)^7 \cdot \frac{1}{b^6}$

${ }^{13} C_2 \frac{a^{11}}{b^2}={ }^{13} C_6 \frac{a^7}{b^6}$

$a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2}=22$

If the coefficient of $x ^7$ 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 $a^4 b^4$ is equal to :

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