If the coefficient of x^7 in (ax - frac{1}{bx | vedclass

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(B) The general term in the expansion of $(ax - \frac{1}{bx^2})^{13}$ is $T_{r+1} = {}^{13}C_r (ax)^{13-r} (-\frac{1}{bx^2})^r = {}^{13}C_r a^{13-r} (

Vedclass · Accepted Answer

$22$

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(B) The general term in the expansion of $(ax - \frac{1}{bx^2})^{13}$ is $T_{r+1} = {}^{13}C_r (ax)^{13-r} (-\frac{1}{bx^2})^r = {}^{13}C_r a^{13-r} (-b^{-1})^r x^{13-3r}$.For the coefficient of $x^7$,set $13-3r = 7$,which gives $r = 2$.Thus,the coefficient of $x^7$ is ${}^{13}C_2 a^{11} b^{-2}$.For the expansion $(ax + \frac{1}{bx^2})^{13}$,the general term is $T_{r+1} = {}^{13}C_r (ax)^{13-r} (\frac{1}{bx^2})^r = {}^{13}C_r a^{13-r} b^{-r} x^{13-3r}$.For the coefficient of $x^{-5}$,set $13-3r = -5$,which gives $r = 6$.Thus,the coefficient of $x^{-5}$ is ${}^{13}C_6 a^7 b^{-6}$.Equating the two coefficients: ${}^{13}C_2 a^{11} b^{-2} = {}^{13}C_6 a^7 b^{-6}$.Dividing both sides by $a^7 b^{-6}$,we get $a^4 b^4 = \frac{{}^{13}C_6}{{}^{13}C_2}$.Calculating the values: ${}^{13}C_6 = \frac{13 	imes 12 	imes 11 	imes 10 	imes 9 	imes 8}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 1716$ and ${}^{13}C_2 = \frac{13 	imes 12}{2 	imes 1} = 78$.Therefore,$a^4 b^4 = \frac{1716}{78} = 22$.

If the coefficient of $x^7$ in $(ax - \frac{1}{bx^2})^{13}$ and the coefficient of $x^{-5}$ in $(ax + \frac{1}{bx^2})^{13}$ are equal,then $a^4 b^4$ is equal to :

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