Question

If ${(a+bx)}^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right)x+...$, then the ordered pair $(a,b)$ equals

• A. $(3,–27)$
• B. $\left(1,\frac{1}{3}\right)$
• C. $(3,9)$
• D. $(3,–9)$

Answer 1

The correct option is D

$(3,–9)$

Explanation for the correct option:

Step 1. Find the value of ordered pair $(a,b)$:

Given, ${(a+bx)}^{-3}=\frac{1}{27}+\left(\frac{1}{3}\right)x+...$ …….(1)

Using the binomial expansion ${\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\mathbf{)}}^{\mathbf{-}\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathit{n}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{!}}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{3}\mathbf{!}}{\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\infty }$,

We have,

$\begin{array}{rcl}{(a+bx)}^{-3}& =& a^{-3}\left[1-3\frac{b}{a}x+6{\left(\frac{bx}{a}\right)}^2+......\infty \right]\\ & =& a^{-3}\left[1-3\frac{b}{a}x+6\frac{b^2{x}^2}{a^2}+......\infty \right]\\ & =& \frac{1}{a^3}-3\frac{b}{a^4}x+6\frac{b^2{x}^2}{a^5}+......\infty ........\left(2\right)\end{array}$

Step 2. Comparing the coefficients of equation(1) and (2), we get

$\frac{1}{a^3}=\frac{1}{27},\frac{-3b}{a^4}=\frac{1}{3}$

$\Rightarrow$$a=3,-3b=\frac{3^4}{3}$

$\Rightarrow$$a=3,b=-9$

$\mathbf{\therefore }\left(a,b\right)\mathbf{}\mathbf{=}\mathbf{}\left(3,-9\right)$

Hence, Option ‘D’ is Correct.