Question

For $\left|x\right|<1,$ the constant term in the expansion of $\frac{1}{{(x-1)}^2(x-2)}$ is

• A. $2$
• B. $1$
• C. $0$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D

$-\frac{1}{2}$

The explanation for the correct option:

Step 1. Construction:

We have$\frac{1}{{(x-1)}^2\left(x-2\right)}$ $={(x-1)}^{-2}{(x-2)}^{-1}$

$={(-1)}^{-2}{(1-x)}^{-2}{(-2)}^{-1}{(1-\frac{x}{2})}^{-1}$

$=\frac{1}{-2}\left[{(1-x)}^{-2}{(1-\frac{x}{2})}^{-1}\right]$

Step 2. Using binomial expansion to find the constant term in the expansion:

We know that

$\begin{array}{rcl}{(1-x)}^{-2}& =& 1+2x+3x^2+4x^3+........\end{array}$ and

${\left(1-\frac{x}{2}\right)}^{-1}=1+\frac{x}{2}+\frac{x}{4}+.....$

So,$\frac{1}{{(x-1)}^2(x-2)}$$\begin{array}{rcl}& =& \frac{-1}{2}[1+2x+3x^2+4x^3+........]\left[1+\frac{x}{2}+{\frac{x}{4}}^2+{\frac{x}{8}}^3+{\frac{x}{16}}^4+.......\right]\end{array}$

$\therefore$Constant term $=\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}$

Hence, option ‘D’ is Correct.