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If $|x| < \frac{1}{2}$, then the coefficient of $x^r$ in the expansion of $\frac{1+2 x}{(1-2 x)^2}$, is
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Verified Answer
The correct answer is:
$(2 r+1) 2^r$
$\begin{aligned}
& \frac{1+2 x}{(1-2 x)^2}=(1+2 x)(1-2 x)^{-2} \\
&=(1+2 x)\left(1+\frac{2}{1 !}(2 x)+\frac{23}{2 !}+\ldots+\frac{23 \ldots r}{(r-1) !}(2 x)^{r-1}\right. \\
&\left.+\frac{2 \cdot 3 \cdot 4 \ldots(r+1)(2 x)^r}{r !}\right)
\end{aligned}$
The coefficient of $x^r$
$\begin{aligned}
& =2 \frac{r !}{(r-1) !} 2^{r-1}+\frac{(r+1) !}{r !} 2^r \\
& =r 2^r+(r+1) 2^r=2^r(2 r+1)
\end{aligned}$
& \frac{1+2 x}{(1-2 x)^2}=(1+2 x)(1-2 x)^{-2} \\
&=(1+2 x)\left(1+\frac{2}{1 !}(2 x)+\frac{23}{2 !}+\ldots+\frac{23 \ldots r}{(r-1) !}(2 x)^{r-1}\right. \\
&\left.+\frac{2 \cdot 3 \cdot 4 \ldots(r+1)(2 x)^r}{r !}\right)
\end{aligned}$
The coefficient of $x^r$
$\begin{aligned}
& =2 \frac{r !}{(r-1) !} 2^{r-1}+\frac{(r+1) !}{r !} 2^r \\
& =r 2^r+(r+1) 2^r=2^r(2 r+1)
\end{aligned}$
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