Question

if $\left|x\right|<\frac{1}{2},$then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{{\left(1-2x\right)}^2}$ is.

• A. $r{2}^r$
• B. $\left(2r-1\right)2^r$
• C. $r{2}^{2r+1}$
• D. $\left(2r+1\right)2^r$

Answer 1

The correct option is D

$\left(2r+1\right)2^r$

Explanation for the correct answer:

Find the value of $x^r$:

Given, $\frac{1+2x}{{\left(1-2x\right)}^2}$

We know the general formula for the expansion ${\left(1+x\right)}^n$is

$T_{r+1}=\frac{n\left(n-1\right).....\left(n-r+1\right)}{r!}{x}^r$

So, the given equation can be written as,

$\begin{array}{rcl}\frac{1+2x}{{\left(1-2x\right)}^2}& =& \left(1+2x\right){\left(1-2x\right)}^{-2}\\ & =& \left(1+2x\right)\left(1+-2\left(-2x\right)\right)+\frac{-2\left(-2-1\right)}{2!}{\left(-2x\right)}^2+....+\frac{-2\left(-2-1\right)\left(-2-2\right).....\left(-2-\left(r-1\right)+1\right)}{\left(r-1\right)!}{\left(-2x\right)}^{r-1}+\frac{-2\left(-2-1\right)\left(-2-2\right).....\left(-2-\left(r\right)+1\right)}{r!}{\left(-2x\right)}^r\\ & =& 1\left(\frac{-2\left(-2-1\right)\left(-2-2\right).....\left(-2-\left(\mathrm{r}\right)+1\right)}{\mathrm{r}!}{\left(-2\mathrm{x}\right)}^{\mathrm{r}}\right)+2\mathrm{x}\left(\frac{-2\left(-2-1\right)\left(-2-2\right).....\left(-2-\left(\mathrm{r}-1\right)+1\right)}{\left(\mathrm{r}-1\right)!}{\left(-2\mathrm{x}\right)}^{\mathrm{r}-1}\right)\\ & =& {\mathrm{x}}^{\mathrm{r}}\left(\left(\mathrm{r}+1\right)2^{\mathrm{r}}+2.\mathrm{r}.2^{\mathrm{r}-1}\right)\mathbf{[}\mathbf{\because }\mathbf{r}\mathbf{!}\mathbf{=}\mathbf{r}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{]}\\ & =& {\mathrm{x}}^{\mathrm{r}}\left(\left(\mathrm{r}+1\right)2^{\mathrm{r}}+\mathrm{r}.2^{\mathrm{r}}\right)\\ & =& {\mathrm{x}}^{\mathrm{r}}\left(\left(2\mathrm{r}+1\right).2^r\right)\end{array}$

Hence, the correct option is D.