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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$\begingroup$Elaborating on J.M. answer, you can procede this way:
$$(1+x)^n = 1+ n x + \frac{1}{2}(n-1) n x^2 + \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$, more generally, $$(1+x)^\alpha = 1 + \sum_{k = 1}^{n} \binom{\alpha}{k} x^k + o(x^n)$$ for any real $\alpha$, where $\binom{\alpha}{k} = \frac{(\alpha -1)(\alpha-2)\ldots (\alpha - k +1)}{k!}$ is the binomial coefficient extended to any real number. Then, you get that $$(1-x)^n = 1 - nx + \frac{1}{2}(n-1)n x^2 - \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$
then, you only have to plug in $n = -1$ to get $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$$
$\endgroup$ $\begingroup$The binomial series works for any real $n$.
$\endgroup$ 2 $\begingroup$The expression $\displaystyle \sum_{k=0}^{\infty} \frac{f^k(0)}{k!} x^k$ is not just the Maclaurin series for functions of the form $f(x)=(1+x)^n$; it is a formula for the Maclaurin series of any function infinitely differentiable at $0$. It gives us the power series in $x$ with all derivatives at $x=0$ matching the derivatives of $f(x)$ at $x=0$.
If you want to use this formula to find the Maclaurin series for $f(x)=\frac{1}{1-x}$, you need to compute all of the derivatives of $f(x)$. --Fortunately, there is a simple pattern you can observe and prove (by induction).
Once you know the derivatives, you can plug in $x=0$, simplify your summation formula, and you'll have your answer.
$\endgroup$ $\begingroup$Try division?
$\dfrac {1}{1-x} = 1 + x + x^2 + x^3 ...$
(edit: I'm not sure why you would try to derive it from the series for $(1+x)^n$ )
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