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Is there a way to derive the Maclaurin series for $\frac{1}{(1-x)}$ after finding the Maclaurin series for $(1+x)^n$ which is $\displaystyle\sum\limits_{k=0}^\infty \frac{f^k(0)}{k!}*x^k$.

From the original equation I could substitute $-1$ for $n$ and $-x$ for $x$ but I don't see how I could do that in the summation. The intention of this is getting the next series without having to do the expansion from scratch.

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    In my opinion, neither of the two responses given thus far directly addresses the question. For that matter, I don't see how to answer it either, and I *really* don't see what the point of the question is: there is no lack of good ways to derive the Maclaurin series for $\frac{1}{1-x}$. My sympathies to the questioner.2010-09-15
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    Pete: I was hoping he'd be able to figure out how to derive the binomial series first, and then make the appropriate substitutions.2010-09-15
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    @J.M.: What you're giving is a common generalization of the two series. As I understand the problem, that doesn't answer it. Of course what you're doing is more general and more useful than what the question wants, but still...2010-09-15
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    Looking at Andy's answer, I think I understand J.M.'s better: you two are both saying that the "$n$" in $(1+x)^n$ is not intended to be a positive integer, i.e., the question really is how to derive the Maclaurin series for $\frac{1}{1-x}$ from the binomial series. *Mea culpa*: I wasn't reading the question that way. Moral: don't use $n$ for a variable that is allowed to take real (or complex...) values. Some people will be confused!2010-09-15
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    No worries Pete, I have to confess I had to overcome my usual convention that $n$ is a variable reserved for nonnegative integers. It was pretty hard, too. :)2010-09-15
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    @Pete: I totally understand and, as a matter of facts, when I wrote $n$ I meant a natural number, then I generalized to $\alpha \in \mathbb{R}$. Then, I addressed the problem with the same notation as the OP (I know, my head's gotta be messed up!)2010-09-15

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