product $abababab...$ in clifford algebra Let $a,b$ are vectors in vector space $V \leq \mathcal{Cl}_n(V)$. 
I would like to know if product $ababab...ab=(ab)^r$ can be written in form $\sum_{\alpha \in A} F_\alpha(a) G_\alpha(b)$. For some $F_\alpha,G_\alpha$ and some index set $A$.
So more precisely:
$(\forall r \in \mathbb{N}) \,\,(\exists \text{set} A) (\exists F_\alpha,G_\alpha : V \rightarrow \mathcal{Cl}_n(V)) (\forall a,b \in V) :abab\dots ab=(ab)^r = \sum_{\alpha \in A} F_\alpha(a) G_\alpha(b)$
Motivation: I'm reading book "geometric algebra to geometric calculus". There they stated generalized Cauchy integral formula and they said that series expansions can be obtained from it in the same way as in complex analysis. But I'm having hard times doing that.
For the start I'm trying to find expansion for $\frac{1}{x-x'}$
$$\frac{1}{x-x'} = \frac{1}{x}\frac{1}{1-x^{-1}x'} = x^{-1} \sum_{i=0}^\infty (x^{-1}x')^i$$
and now I need to write $x^{-1} \sum_{i=0}^\infty (x^{-1}x')^i$ in form $\sum_\alpha F_\alpha(x) G_\alpha(x')$ to be of any use.
In paper "Quaternionic analysis" by A. Sudbery. There is proposition 10 which says something similar for quaternions. Unfortunately proof is given only briefly and I do not understand it.
 A: I can give an answer in a particular case $a.b=-b.a$.
You have to use the identity $a.b.a.b= - a.a.b.b= -(a.b)^2$. Say the product $a.b=v$. 
Then for  $n=2$, $a.b.a.b= - v^2$.
For $n=3$,  $a.b.a.b.a.b =- a.b.a.a.b.b = a.a.b.a.b.b = - a.a.a.b.b.b = - v^3$ .
For $n=4$, $a.b.a.b.a.b.a.b = a . a . a . a . b . b . b . b= v^4$.
For $n=5$, $ a.b.a.b.a.b.a.b.a.b = a . a . a . a . a . b . b . b . b . b= v^5$.
For $n=6$, $a.b.a.b.a.b.a.b.a.b.a.b =-a . a . a . a . a . a . b . b . b . b . b . b = -v^6$
For $n=7$, $a.b.a.b.a.b.a.b.a.b.a.b.a.b = -a . a . a . a . a . a . a . b . b . b . b . b . b . b =- v^7$
For $n=8$,  $a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b = a . a . a . a . a . a . a . a . b . b . b . b . b . b . b . b = v^8$.
For $n=9$,  $a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b= a . a . a . a . a . a . a . a . a . b . b . b . b . b . b . b . b . b = v^9$ .
For $n=10$, $ a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b.a.b =-a . a . a . a . a . a . a . a . a . a . b . b . b . b . b . b . b . b . b . b = - v^{10}$.
So the sign pattern is $ [1,-1,-1,1,1,-1,-1,1,1,-1, \ldots]$.
Therefore, the formula can be calculated as $-v^n$ if $n-2 \mod 4 =\{0,1\}$ or $v^n$ if $n-2 \mod 4 =\{2,3\}$.
A: Let $x$ and $y$ be two variables, let $k$ be a field and let $t\in k$. Consider the algebra $A_t=k\langle x,y\rangle/(xy+yx-t)$.
It is very easy to see that the set of all elements in $A$ of the form $x^iy^j$ with $i$, $j\geq0$ is a basis of $A_t$ as a vector space. This means that, in particular, if we fix $n\geq0$ there exists scalars $\alpha_{i,j}$ for all $i$, $j\geq0$, almost all of which are zero, such that in $A_t$ we have $$(xy)^n=\sum_{i,j\geq0}\alpha_{i,j}x^iy^j. \tag{$\star$}$$ The coefficients depend on $n$ and on $t$, clearly.
Now let $a$ and $b$ be elements in your vector space $V$, and let $t$ be such  that in the $Cl(V)$ we have $ab+ba=t$. Then there is exactly one homomorphism of algebras $\phi:A_t\to Cl(V)$ such that $\phi(x)=a$ and $\phi(y)=b$, and then we have in $Cl(V)$ that $$(ab)^n=\sum_{i,j\geq0}\alpha_{i,j}a^ib^j.$$ Indeed,this simply follows from applying the map $\phi$ to the previous equality.
The answer to your question is thus yes.
Computing with a computer, I find that the formulas above for the first few values of $n$ are
$$
\begin{array}{l}
 (x y)^1\to x y \\
 (x y)^2\to t x y-x^2 y^2 \\
 (x y)^3\to t^2 x y-t x^2 y^2-x^3 y^3 \\
 (x y)^4\to t^3 x y-t^2 x^2 y^2-2 t x^3 y^3+x^4 y^4 \\
 (x y)^5\to t^4 x y-t^3 x^2 y^2-3 t^2 x^3 y^3+2 t x^4 y^4+x^5 y^5 \\
 (x y)^6\to t^5 x y-t^4 x^2 y^2-4 t^3 x^3 y^3+3 t^2 x^4 y^4+3 t x^5 y^5-x^6 y^6 \\
 (x y)^7\to t^6 x y-t^5 x^2 y^2-5 t^4 x^3 y^3+4 t^3 x^4 y^4+6 t^2 x^5 y^5-3 t x^6 y^6-x^7 y^7 \\
 (x y)^8\to t^7 x y-t^6 x^2 y^2-6 t^5 x^3 y^3+5 t^4 x^4 y^4+10 t^3 x^5 y^5-6 t^2 x^6 y^6-4 t x^7
   y^7+x^8 y^8
\end{array}
$$
This seems to be
$$
\sum _{k=0}^{n-1} (-1)^{\frac{1}{2} k (k+1)} \binom{n-1-\left\lfloor \tfrac{k+1}{2}\right\rfloor
   }{\left\lfloor \tfrac{k}{2}\right\rfloor } t^{n-1-k} x^{k+1} y^{k+1}
$$
and an induction should easily prove this.
