The coefficient of x^{-7} in the expansion of | vedclass

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(A) The general term in the expansion of $(ax - \frac{1}{bx^2})^{11}$ is given by $T_{r+1} = ^{11}C_r (ax)^{11-r} (-\frac{1}{bx^2})^r$.$T_{r+1} = ^{11

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$\frac{462a^5}{b^6}$

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(A) The general term in the expansion of $(ax - \frac{1}{bx^2})^{11}$ is given by $T_{r+1} = ^{11}C_r (ax)^{11-r} (-\frac{1}{bx^2})^r$.$T_{r+1} = ^{11}C_r a^{11-r} x^{11-r} (-1)^r b^{-r} x^{-2r}$$T_{r+1} = ^{11}C_r a^{11-r} (-1)^r b^{-r} x^{11-3r}$To find the coefficient of $x^{-7}$,we set the exponent of $x$ equal to $-7$:$11 - 3r = -7$$3r = 18 \Rightarrow r = 6$Substituting $r = 6$ into the expression for the coefficient:Coefficient $= ^{11}C_6 a^{11-6} (-1)^6 b^{-6}$$= ^{11}C_6 a^5 \frac{1}{b^6}$Since $^{11}C_6 = ^{11}C_5 = \frac{11 	imes 10 	imes 9 	imes 8 	imes 7}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 462$Therefore,the coefficient is $\frac{462a^5}{b^6}$.

The coefficient of $x^{-7}$ in the expansion of $\left( ax - \frac{1}{bx^2} \right)^{11}$ is

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