Question

If the coeffcient of $x^8$ in $\left(ax^2+\frac{1}{bx}\right)^{13}$is equal to the coefficient of $x^{-8}$ in $\left(ax-\frac{1}{bx^2}\right)^{13}$ then $a$ and $b$ will satisfy the relation.

• A. ab + 1 = 0
• B. ab = 1
• C. a= 1 - b
• D. a + b = - 1

Answer 1

Answer: (A)

$In \left(ax^{2}+\frac{1}{bx}\right)^{13}$, $T_{r-1} = \,{}^{13}C_{r}\left(ax^{2}\right)^{13-r} \left(\frac{1}{bx}\right)^{r}$
$= \,{}^{13}C_{r} \,a^{13} \,{}^{r}x^{26} \,{}^{3r}$
For co-eff of $x^{8} = 26 - 3r =8$
$\Rightarrow 3r$
$\Rightarrow 18 \,r = 6$
$\therefore $ co-eff of $x^{8} = \,{}^{13}C_{6}\,a^{7}\cdot\frac{1}{b^{6}}$
Again for $\left(ax-\frac{1}{bx}\right)^{13}$, $T_{r+1} = \,{}^{13}C_{r}\,\left(ax\right)^{13-r}\,\left(-\frac{1}{bx^{2}}\right)$
$= \,{}^{13}C_{r}\,a^{13}\,{}^{r}\,x^{13}\,{}^{r}\,{}^{2r} \,\left(\frac{1}{b}\right)^{r}$
$= \,{}^{13}C_{r}\,a^{13-r}\,x^{13-3r} \,\left(- \frac{1}{b}\right)^{r}$
For co-eff. of $x^{-8}$, $\,13 - 3r = - 8$
$\Rightarrow 21 = 3r$
$\Rightarrow r= 7$
$\therefore $ co-eff. of $x^{-8}$ is $^{13}C_{7} \,a^{6} \left(\frac{1}{b}\right)^{7} = \,{}^{13}C_{6}\,\left(-\frac{a^{6}}{b^{7}}\right)$
By the given condition
$^{13}C_{6} \frac{a^{7}}{b^{6}} = -\,{}^{13}C_{6} \frac{a^{6}}{b^{7}}$
$\Rightarrow a = - \frac{1}{b}$
$\Rightarrow ab + 1 = 0$