Let a and b be two nonzero real numbers. If t | vedclass

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(B) The general term in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is $T_{r+1} = {}^4C_r (ax^2)^{4-r} (\frac{70}{27bx})^r = {}^4C_r a^{4-r} (\frac{

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$3$

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(B) The general term in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is $T_{r+1} = {}^4C_r (ax^2)^{4-r} (\frac{70}{27bx})^r = {}^4C_r a^{4-r} (\frac{70}{27b})^r x^{8-3r}$.For the coefficient of $x^5$,we set $8-3r = 5$,which gives $r=1$.The coefficient is ${}^4C_1 a^3 (\frac{70}{27b}) = 4a^3 \cdot \frac{70}{27b} = \frac{280a^3}{27b}$.The general term in the expansion of $(ax - \frac{1}{bx^2})^7$ is $T_{r+1} = {}^7C_r (ax)^{7-r} (-\frac{1}{bx^2})^r = {}^7C_r a^{7-r} (-\frac{1}{b})^r x^{7-3r}$.For the coefficient of $x^{-5}$,we set $7-3r = -5$,which gives $3r = 12$,so $r=4$.The coefficient is ${}^7C_4 a^3 (-\frac{1}{b})^4 = 35 \cdot \frac{a^3}{b^4} = \frac{35a^3}{b^4}$.Equating the two coefficients: $\frac{35a^3}{b^4} = \frac{280a^3}{27b}$.Dividing by $35a^3$ (since $a 
eq 0$),we get $\frac{1}{b^3} = \frac{8}{27}$.Thus,$b^3 = \frac{27}{8}$,which implies $b = \frac{3}{2}$.Therefore,$2b = 2(\frac{3}{2}) = 3$.

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is equal to the coefficient of $x^{-5}$ in the expansion of $(ax - \frac{1}{bx^2})^7$,then the value of $2b$ is

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