If the coefficients of x^7 in left(ax^2+frac{ | vedclass

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(B) For the expansion $\left(ax^2+\frac{1}{2bx}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax^2)^{11-r} (\frac{1}{2bx})^r = {}^{11}C_r a^

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$729ab = 32$

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(B) For the expansion $\left(ax^2+\frac{1}{2bx}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax^2)^{11-r} (\frac{1}{2bx})^r = {}^{11}C_r a^{11-r} (\frac{1}{2b})^r x^{22-3r}$.Setting $22-3r = 7$,we get $3r = 15$,so $r = 5$. The coefficient is ${}^{11}C_5 a^6 (\frac{1}{2b})^5 = \frac{{}^{11}C_5 a^6}{32b^5}$.For the expansion $\left(ax-\frac{1}{3bx^2}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax)^{11-r} (-\frac{1}{3bx^2})^r = {}^{11}C_r a^{11-r} (-\frac{1}{3b})^r x^{11-3r}$.Setting $11-3r = -7$,we get $3r = 18$,so $r = 6$. The coefficient is ${}^{11}C_6 a^5 (-\frac{1}{3b})^6 = \frac{{}^{11}C_6 a^5}{729b^6}$.Equating the coefficients: $\frac{{}^{11}C_5 a^6}{32b^5} = \frac{{}^{11}C_6 a^5}{729b^6}$.Since ${}^{11}C_5 = {}^{11}C_6 = 462$,we have $\frac{a^6}{32b^5} = \frac{a^5}{729b^6}$.Dividing by $a^5$ and multiplying by $b^6$,we get $\frac{ab}{32} = \frac{1}{729}$,which implies $729ab = 32$.

If the coefficients of $x^7$ in $\left(ax^2+\frac{1}{2bx}\right)^{11}$ and $x^{-7}$ in $\left(ax-\frac{1}{3bx^2}\right)^{11}$ are equal,then

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