Need help to further simplify the expression $\sum_{i = 1}^{6} (I - AX)^i $ I am working on one research problem where I need help to simplify the following algebraic expression 
$$\sum_{i = 1}^6 (I - AX)^i = (I - AX) + (I - AX)^2 + (I -
AX)^3+ \cdots + (I - AX)^6$$
where $A$ and $X$ are matrices of order $n \times n$. 
Edit: $I$ is identity matrix of order $n\times n$
I am not able to find any simpler way to further simplify this expression. Please help me in this regard. Thank you.
 A: $\mathbf I$ commutes with any matrix, so you can always write
$$
\begin{gathered}
  \sum\limits_{i = 1}^6 {\left( {\mathbf{I - AX}} \right)^{\,i} }  = \sum\limits_{i = 1}^6 {\sum\limits_{k = 0}^{\left( i \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  i \\ 
  k \\ 
\end{gathered}  \right)\left( {\mathbf{AX}} \right)^{\,k} } }  =  \hfill \\
   = \sum\limits_{0 \leqslant k}^{} {\sum\limits_{i = 1}^6 {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  i \\ 
  k \\ 
\end{gathered}  \right)\left( {\mathbf{AX}} \right)^{\,k} } }  = \sum\limits_{0 \leqslant k}^{\left( 6 \right)} {\left( { - 1} \right)^{\,k} \left( {\left( \begin{gathered}
  7 \\ 
  k + 1 \\ 
\end{gathered}  \right) - \left( \begin{gathered}
  1 \\ 
  k + 1 \\ 
\end{gathered}  \right)} \right)\left( {\mathbf{AX}} \right)^{\,k} }  =  \hfill \\
   = 6\,\mathbf{I} + \sum\limits_{1 \leqslant k}^{\left( 6 \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  7 \\ 
  k + 1 \\ 
\end{gathered}  \right)\left( {\mathbf{AX}} \right)^{\,k} }  \hfill \\ 
\end{gathered} 
$$
Then, if $\mathbf A$ and $\mathbf X$ do not commute, you cannot apply the distributive property of the power index
with respect of the product, e.g.
$$
\left( {\mathbf{AX}} \right)^{\,2}  = \mathbf{AXAX} \ne \mathbf{A}^{\,2} \mathbf{X}^{\,2} 
$$
Note:
If, as you say in the comment , $\mathbf X$ is the pseudoinverse of $\mathbf A$,
then
$$
\mathbf{AX} = \mathbf{AA}^{\,\mathbf{ + }} 
$$
shall be a Projection Matrix as well as $\mathbf {(I-AX)}$, thus they shall be idempotent, i.e.:
$$
\left( {\mathbf{AX}} \right)^{\,2}  = \mathbf{AX}\quad \left( {\mathbf{I - AX}} \right)^{\,2}  = \mathbf{I - AX}
$$
and then the simplification of the sum is obvious.
A: Let $\mathrm Y := \mathrm A \mathrm X$. Using SymPy Live:
>>> y = symbols('y',real=True)
>>> expand((1-y) + (1-y)**2 + (1-y)**3 + (1-y)**4 + (1-y)**5 + (1-y)**6)
 6      5       4       3       2           
y  - 7*y  + 21*y  - 35*y  + 35*y  - 21*y + 6

or, alternatively,
>>> y = symbols('y',real=True)
>>> expand(sum([(1-y)**k for k in range(1,7)]))
 6      5       4       3       2           
y  - 7*y  + 21*y  - 35*y  + 35*y  - 21*y + 6

Since $\mathrm I$ and $\mathrm Y$ commute,
$$\sum_{k=1}^6 (\mathrm I - \mathrm Y)^k = 6 \mathrm I - 21 \mathrm Y + 35 \mathrm Y^2 - 35 \mathrm Y^3 + 21 \mathrm Y^4 - 7 \mathrm Y^5 + \mathrm Y^6$$
