Maclaurin series expansion of $\frac{1}{(1+x)^n}$ I am trying to figure out the Maclaurin Series expansion of the function, preferribly in a sneaky and clever way. Any ideas?
Thanks.
 A: set
$$
F(x)=\frac1{(1-x)^n} = \prod_{j=1}^n\sum_{k=0}^{\infty}x^k
$$
here the coefficient of $x^m$ is the number of ways of expressing $m$ as a sum of $n$ non-negative integers, with order significant (i.e. $3+4+5$ and $4+3+5$ are different expressions).
this is just the number of ways of fitting $n-1$ separators into $m$ slots, i.e. $\binom{n+m-1}{m}$
so:
$$
\frac1{(1+x)^n}=F(-x)=1-nx+\frac{n(n+1)}{2!}x^2-...
$$
A: It's almost the $n^{\rm th}$ derivative of $1/(1+x)$. 
A: I think that computing the derivatives is really the easiest way (as pointed out By Andre Nicolas). Your function is 
$$
f(x) = (1+x)^{-n}\,.
$$
Now differentiate:
\begin{eqnarray}
f'(x) &=& -n (1+x)^{-(n+1)} \\
f''(x) &=& (-1)^2 n(n+1) (1+x)^{-(n+2)}\,.
\end{eqnarray}
By now the general pattern should be clear:
$$
f^{(k)} (x) = (-1)^k \, n (n+1) \cdots (n+k-1) \, (1+x)^{-n+k}.
$$
Evaluating the above at $x=0$ you get the following Maclaurin series (convergent for $|x|<1$)
$$
f(x) = \sum_{k=0}^{\infty} a_k\, x^k$$
with 
$$
a_k = \frac{f^{(k)}(0)}{k!} = (-1)^k \frac{ n (n+1) \cdots (n+k-1)}{k!} = (-1)^k \left ( \begin{array}{c} n+k-1\\ k \end{array} \right ).
$$
A: Simple, yet not straightforward, approach:
\begin{align}
\frac{1}{(1-x)^n} &= (1-x)^{-n} \\
&= \sum_{k=0}^\infty\binom{-n}{k}(-x)^k\tag 1 \\
&= \sum_{k=0}^\infty\left((-1)^k\binom{n+k-1}{k}\right)(-x)^k\tag2 \\
&= \sum_{k=0}^\infty\binom{n+k-1}{k}x^k \\
\end{align}
We arrive at $(1)$ using the binomial series, and at $(2)$ using the relationship between the multiset coefficient and the binomial coefficient $\binom{-n}{k}$.
