Question

If the coefficient 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 $a\&b$ 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

$ab+1=0$

Explanation for the correct option.

Step 1: Write the general term

We know that the general term in ${\left(u+v\right)}^n={}^{n}C_r{u}^{n-r}{v}^r$.

Therefore, the general term of ${\left(a{x}^2+\frac{1}{bx}\right)}^{13}$ can be written as:

Here, $n=13,u=a{x}^2,v=\frac{1}{bx}$.

$\begin{array}{rcl}{\mathit{T}}_{\mathbf{r}\mathbf{+}\mathbf{1}}& \mathbf{=}& {}^{\mathbf{13}}\mathit{C}_{\mathbf{r}}\mathbf{·}{\mathbf{\left(}\mathbf{a}{\mathbf{x}}^{\mathbf{2}}\mathbf{\right)}}^{\mathbf{(}\mathbf{13}\mathbf{-}\mathbf{r}\mathbf{)}}\mathbf{·}\frac{\mathbf{1}}{\mathbf{(}\mathbf{b}\mathbf{x}{\mathbf{)}}^{\mathit{r}}}\end{array}$

This term can also be written as,

$\begin{array}{rcl}{T}_{r+1}& =& {}^{13}\mathrm{C}_{\mathrm{r}}·x^{26-3\mathrm{r}}\frac{{\mathrm{a}}^{13-\mathrm{r}}}{{\mathrm{b}}^{\mathrm{r}}}\end{array}$

Step 2: Find the coefficient of $x^8\&x^{-8}$

For the power of $\mathrm{x}$ is $8$:
$\begin{array}{rcl}& \Rightarrow & 26-3\mathrm{r}=8\\ \mathrm{r}& =& 6\end{array}$

For the power of $x$ to be $-8$.
$\begin{array}{rcl}& \Rightarrow & 13-3\mathrm{r}=-8\\ \mathrm{r}& =& 7\end{array}$

Step 3: Find the relation between $a\&b$

Now according to the question:
$\begin{array}{rcl}\therefore {}^{13}\mathrm{C}_6·\frac{{\mathrm{a}}^7}{{\mathrm{b}}^6}& =& -{}^{13}\mathrm{C}_7·\frac{{\mathrm{a}}^6}{{\mathrm{b}}^7}\dots .\left({}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{x}}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}-\mathrm{x}}\right)\\ & \Rightarrow & \mathrm{ab}=-1\\ & \Rightarrow & ab+1=0\end{array}$

Hence, option (A) is correct.