Question

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

The correct option is A $ab+1=0$

The genral term of ${(a{x}^2+\frac{1}{bx})}^{13}$ can be written as $T_{r+1}$=${}^{13}{C}_r.{\left(a{x}^2\right)}^{(13-r)}.\frac{1}{{\left(bx\right)}^r}$=${}^{13}{C}_r.x^{26-3r}\frac{a^{13-r}}{b^r}$

$T_{r+1}^{\prime }$ be the term for which power of $x$ is $8$

$\Rightarrow$ $26-3r=8$

$\Rightarrow$ $r=6$

Coefficent of $T_{r+1}^{\prime }$=${}^{13}{C}_6.\frac{a^7}{b^6}$

The genral term of ${(ax-\frac{1}{b{x}^2})}^{13}$ can be written as

$T_{r+1}$=${}^{13}{C}_r.{\left(ax\right)}^{13-r}.\frac{1}{{\left(b{x}^2\right)}^r}.{(-1)}^r$=${}^{13}{C}_r.x^{13-3r}.\frac{a^{13-r}}{b^r}{(-1)}^r$

$T{"}_{r+1}$ be the term for which power be -8

$\Rightarrow$ $13-3r=-8$

$\Rightarrow$ $r=7$

Coefficent of $T{"}_{r+1}$=${-}^{13}{C}_7.\frac{a^6}{b^7}$

As coefficent of $T_{r+1}^{\prime }$ =$T{"}_{r+1}$

$\Rightarrow$${}^{13}{C}_6.\frac{a^7}{b^6}={-}^{13}{C}_7.\frac{a^6}{b^7}$ ....As ${}^n{C}_x{=}^n{C}_{n-x}$

$\Rightarrow$$ab=-1$

$\Rightarrow$$ab+1=0$