Q. If $|x| < \frac{1}{2}$, then the coefficient of $x^{r}$ in the expansion of $\frac{1 + 2x}{{(1 - 2x)}^{2}}$, is
A) $r2^{r}$
B) $(2r - 1)2^{r}$
C) $r2^{2r + 1}$
D) $(2r + 1)2^{r}$
Explanation
(d)
$\frac{1 + 2x}{{(1 - 2x)}^{2}} = (1 + 2x){(1 - 2x)}^{- 2}$
$= (1 + 2x)\left( 1 + \frac{2}{1!}(2x) + \frac{2 \bullet 3}{2!}(2x)^{2} + ... + \frac{2 \bullet 3\ldots r}{(r - 1)!}(2x)^{r - 1} + \frac{2 \bullet 3 \bullet 4\ldots(r + 1)(2x)^{r}}{r!} \right)$
The coefficient of $x^{r} = 2\frac{r!}{(r - 1)!}2^{r - 1} + \frac{(r - 1)!}{r!}2^{r}$
$= r2^{r} + (r + 1)2^{r} = 2^{r}(2r + 1) = =$ !)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.