If the coefficient of x^5 in the expansion of | vedclass

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

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(A) In the expansion of $(ax^2+\frac{1}{bx})^{13}$,the general term is $T_{r+1} = {}^{13}C_r (ax^2)^{13-r} (\frac{1}{bx})^r = {}^{13}C_r \frac{a^{13-r}}{b^r} x^{26-3r}$.For the coefficient of $x^5$,we set $26-3r = 5$,which gives $3r = 21$,so $r = 7$.The coefficient is ${}^{13}C_7 \frac{a^6}{b^7}$.In the expansion of $(ax-\frac{1}{bx^2})^{13}$,the general term is $T_{r+1} = {}^{13}C_r (ax)^{13-r} (-\frac{1}{bx^2})^r = (-1)^r {}^{13}C_r \frac{a^{13-r}}{b^r} x^{13-3r}$.For the coefficient of $x^{-5}$,we set $13-3r = -5$,which gives $3r = 18$,so $r = 6$.The coefficient is $(-1)^6 {}^{13}C_6 \frac{a^7}{b^6} = {}^{13}C_6 \frac{a^7}{b^6}$.Equating the coefficients: ${}^{13}C_7 \frac{a^6}{b^7} = {}^{13}C_6 \frac{a^7}{b^6}$.Since ${}^{13}C_7 = {}^{13}C_6$,we have $\frac{a^6}{b^7} = \frac{a^7}{b^6}$.Dividing both sides by $a^6$ and multiplying by $b^7$,we get $1 = ab$.

If the coefficient of $x^5$ in the expansion of $(ax^2+\frac{1}{bx})^{13}$ is equal to the coefficient of $x^{-5}$ in the expansion of $(ax-\frac{1}{bx^2})^{13}$,then $ab=$

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