Question

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

$In{\left(a{x}^2+\frac{1}{bx}\right)}^{13}$, $T_{r-1}={}^{13}{C}_r{\left(a{x}^2\right)}^{13-r}{\left(\frac{1}{bx}\right)}^r$
$={}^{13}{C}_r{a}^{13}{}^r{x}^{26}{}^{3r}$
For co-eff of $x^8=26-3r=8$
$\Rightarrow 3r$
$\Rightarrow 18r=6$
$\therefore$ co-eff of $x^8={}^{13}{C}_6{a}^7\cdot \frac{1}{b^6}$
Again for ${\left(ax-\frac{1}{bx}\right)}^{13}$, $T_{r+1}={}^{13}{C}_r{\left(ax\right)}^{13-r}\left(-\frac{1}{b{x}^2}\right)$
$={}^{13}{C}_r{a}^{13}{}^r{x}^{13}{}^r{}^{2r}{\left(\frac{1}{b}\right)}^r$
$={}^{13}{C}_r{a}^{13-r}{x}^{13-3r}{\left(-\frac{1}{b}\right)}^r$
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
${}^{13}{C}_6\frac{a^7}{b^6}=-{}^{13}{C}_6\frac{a^6}{b^7}$
$\Rightarrow a=-\frac{1}{b}$
$\Rightarrow ab+1=0$