Question

Let a and b be two nonzero real numbers. If the coefficient of x5 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

Answer 1

$T_{r+1}=\ ^4C_r(ax^2)^{(4-r)}\times(\frac{70}{27bx})^r$
For coefficient of $x^5$,$8-2r-r=5\Rightarrow r=1$
$\Rightarrow$ Coefficient of $x^5 = \ ^4C_1a^3(\frac{70}{27b})$
$t_{r+1}=^7C_r(ax)^{7-r}(-\frac{1}{bx^2})^r$
for the coefficient of x-5, $7-r-2r=-5\Rightarrow r=4$
Coefficient of x-5 = $^7C_4a^3\frac{1}{b^4}\Rightarrow 2b=3$
The value of 2b is 3.
Therefore, the correct answer is 3.