If the coefficient of x 15 in the expansion of a x 3 + 1 b x 1 3 15 is equal to the coefficient of x - 15 in…
If the coefficient of in the expansion of is equal to the coefficient of in the expansion of , where and are positive real numbers, then for each such ordered pair :
Solution
$(r+1)^{th}$ term in the expansion of $(ax^3 + \frac{1}{bx^{\frac{1}{3}}})^{15}$ is
$T_{r+1} = C_{r}^{15} (ax^3)^{15-r} (\frac{1}{bx^{\frac{1}{3}}})^r$
$\Rightarrow T_{r+1} = C_{r}^{15} a^{15-r} x^{45-10r/3} b^r$
For the coefficient of $x^{15}$ in $(ax^3 + \frac{1}{bx^{\frac{1}{3}}})^{15}$:
$45 - \frac{10r}{3} = 15$
$\Rightarrow 30 = \frac{10r}{3}$
$\Rightarrow r = 9$
Coefficient of $x^{15} = C_{9}^{15} a^6 b^{-9}$
$(r+1)^{th}$ term in the expansion of $(ax^{\frac{1}{3}} - \frac{1}{bx^3})^{15}$ is:
$T_{r+1} = C_{r}^{15} (ax^{\frac{1}{3}})^{15-r} (-\frac{1}{bx^3})^r$
For the coefficient of $x^{-15} in (ax^{\frac{1}{3}} - \frac{1}{bx^3})^{15}$ is:
$5 - \frac{r}{3} - 3r = -15$
$\Rightarrow \frac{10r}{3} = 20$
$\Rightarrow r = 6$
Coefficient of $x^{-15} = C_{6}^{15} a^9 b^{-6}$
Hence,
$C_{9}^{15} a^6 b^{-9} = C_{6}^{15} a^9 b^{-6}$
$\Rightarrow \frac{a^9 b^6}{a^6 b^9} = 1$
$\Rightarrow a^3 b^3 = 1 \Rightarrow ab = 1$