How do I find the MacLaurin series for $\frac{1}{1 - 2x}$?

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SUMMARY

The MacLaurin series for the function $\frac{1}{1 - 2x}$ is derived by substituting $2x$ into the known series for $\frac{1}{1 - x}$, resulting in the series $\sum_{n=0}^{\infty} (2x)^n$. This series converges for $|x| < \frac{1}{2}$. The preferred notation for expressing this series as a MacLaurin series is $\sum_{n=0}^{\infty} c_n x^n$, where $c_n = 2^n$. This notation emphasizes the coefficients associated with each power of $x$.

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I need to find the maclaurin series of the function

$$\frac{1}{1 - 2x}$$.

I know $\frac{1}{1 - x}$ is $1 + x + x^2 + x^3 ...$ but how can I use this to solve the problem? I don't think I can just plug in $2x$ can I?
 
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Yes; plug in $2x$ and note that the series converges for $|x|<\dfrac12$.
 
greg1313 said:
Yes; plug in $2x$ and note that the series converges for $|x|<\dfrac12$.

So, I would get something like

$$\sum_{n = 0}^{\infty} 2^nx^n$$

Is this correct?
 
Yes, but I think the notation

$$\sum_{n=0}^\infty(2x)^n$$

would be preferred. :)
 
greg1313 said:
Yes, but I think the notation

$$\sum_{n=0}^\infty(2x)^n$$

would be preferred. :)

Actually, as it's specifically asked to be written as a MacLaurin Series, where it should be written $\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{c_n\,x^n} \end{align*}$ the OP's notation would be preferred.
 

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