Binomial Theorem Question 12
Question 12 - 30 January - Shift 1
If the coefficient of $x^{15}$ in the expansion of $(ax^{3}+\frac{1}{b x^{\frac{1}{3}}})^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $(a x^{\frac{1}{3}}-\frac{1}{b x^{3}})^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(a, b)$ :
(1) $a=b$
(2) $ab=1$
(3) $a=3 b$
(4) $ab=3$
Show Answer
Answer: (2)
Solution:
Formula: Important Results Coefficient of $x^{m}$ in the expansion of $\left(a x^{p} + \frac{b}{x^{q}}\right)^{n}$
Option (2)
Coefficient Of $x^{15}$ in $(a x^{3}+\frac{1}{b x^{1 / 3}})^{15}$
$ T _{r+1}={ }^{15} C_r(ax^{3})^{15-r}(\frac{1}{bx^{1 / 3}})^{r} $
$45-3 r-\frac{r}{3}=15$
$30=\frac{10 r}{3}$
$r=9$
Coefficient of $x^{15}={ }^{15} C_9 a^{6} b^{-9}$
Coefficient of $x^{-15}$ in $(ax^{1 / 3}-\frac{1}{bx^{3}})^{15}$
$T _{r+1}={ }^{15} C_r(ax^{1 / 3})^{15-r}(-\frac{1}{bx^{3}})^{r}$
$5-\frac{r}{3}-3 r=-15$
$\frac{10 r}{3}=20$
$r=6$
So, $ { }^{15} C_9 a^6 b^{-9}={ }^{15} C_6 a^9 b^{-6} $
$ \Rightarrow a^{-3} b^{-3}=1 $
or $ a b=1 $
