What is the binomial expansion for $(a+b)^{-1}$? What is the binomial expansion for $(1+x)^{-1}$?
I really don't know how to solve it using negative powers.
 A: There is not an expansion for this by the usual binomial expansion. However, there is one by the generalized binomial theorem, which gives, for $n \in \Bbb Z^+$,
$$(1-x)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{n-1} x^k$$
Factor out $a$ from your expression and replace adding $b$ with subtracting $(-b)$:
$$(a+b)^{-1} = a^{-1} \left(1- \left(-\frac b a \right) \right)^{-1}$$
Now apply the theorem for $x=-b/a$ and $n=1$.

Footnote:
You could approach this by the geometric series instead, but this generalizes it somewhat. I used the above solution because "binomial" was specified, but geometric is easier. In the geometric series, we have
$$(1-x)^{-1} = \frac{1}{1-x} = \sum_{k=0}^\infty x^k$$
Do the factoring as before and take $x=-b/a$ for the same result.
A: If $|x| <1$,
$\frac{1}{1+x} = (1+x)^{-1} = 1 -x + x^2 - x^3 \cdots$
A: Geometric series ($q=-x$):
$\frac{1}{1-q} = \sum_{n=0}^\infty q^n.$
A: Using the binomial expansion:
$$ (x + a)^n = C_0^n x^n + C_1^n x^{n-1} a + C_2^n x^{n-2} a^2 . . . . . C_n^n a^n $$
For $|x|<1$, so the series converges. Therefore we can take $n =-n$
You can form these expansion for $ (x+ a) $ as well, now substituting $1$ in place of $a$ and $ -1 $ in place of $n$
The equation is formed as :
$$ (1+x)^{-1} = 1 - x + \frac { 2 x^2} {2!} 
- \frac { 3*2 x^3 } {3!} +. . .  . (-1)^nx^n $$
A: Observe that $(1+x) ^n = C^n_0 + C^n_1 x + \cdots + C^n_n x^n$ for $n\in \mathbb N$, and that
$$C^n_k = \dfrac {n!} {k! (n-k)!} = \dfrac{n(n-1) \cdots (n-k+1)} {k!} $$
Observe that under this definition $C^n_k =0$ for $k>n$ where $n, k\in\mathbb N$ since the numerator contains a zero in the product, so let this binomial sum run for $k\in \mathbb N$, i.e. let the upper limit of $k$ goes to infinity. 
We now generalize this to negative $n$, and thus we have:
$$(1+x) ^{-1} = \sum_{k=0} ^\infty C^{-1} _k x^k$$
But $ C^{-1} _k= \dfrac {(-1) (-2) \cdots (-k)} {k!}  = (-1) ^k$
So we have
$$(1+x) ^{-1} = \sum_{k=0} ^\infty (-1) ^k x^k$$
