Noncommutative rings, finding $a$ and $b$ such that every term in the sum $(a+b)^n = a^n + a^{n-1}b + \ldots$ is distinct This question was inspired by the binomial  theorem for rings. For commutative rings, we have the identity
$$(a+b)^n = \sum_{k=0}^n {n \choose k}a^kb^{n-k}$$
which does not hold for non-commutative rings. However, we can still expand $(a+b)^n$ to get
\begin{align*}
(a+b)^2 &= a^2 + ab + ba + b^2 \\
(a+b)^3 &= a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 \\
\cdots \\
(a+b)^n &= aa\ldots a + aa \ldots ab + aa \ldots ba + aa \ldots bb + \ldots
\end{align*}
This motivates the following question. For every $n$, is it possible to find a ring $R$ with elements $a$ and $b$ such that the $2^n$ terms in the expansion of $(a+b)^n$ are distinct?
There is also a stronger question: is it possible to find $a$ and $b$ such that for every $n$, this is true?
For example, taking $A = \begin{bmatrix}2 &1 \\0 &1\end{bmatrix}$ and $B = \begin{bmatrix}1 &2 \\0 & 1\end{bmatrix}$ from $M_2(\mathbb{Z})$ works for $n = 1, 2, 3$ and $4$ but fails for $n = 5$ because $ABBBA = BBAAB$. 
 A: Take the polynomial ring over a field in two noncommuting variables: $\mathbb{F}\langle x,y\rangle$. I've put them in langle/rangle brackets to remind everyone that they do not commute with each other. 
Then $x+y$ has your property.
Generally any free object on two generators is going to do something like that, because free objects don't satisfy any relations. For example, in commutative rings the relation $xy-yx=0$ allows you to "shuffle" different looking expressions until they are identical (like $aab$ and $aba$).
A: One way to do this is to take $R$ to be the group ring $\mathbb{Z}[F_2]$, where $F_2$ is the free group on generators $a$ and $b$. Then the fact that the summands are all different (for any $n$) is a direct consequence of the freeness of the group.
Of course, one doesn't have to explicitly embed a free group in a ring for a free group to be present. For example, $\begin{pmatrix} 1 & 2 \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 2 & 1\end{pmatrix}$ generate a free group of rank $2$, so you can take $R$ to be the ring of two-by-two integer matrices and these matrices as $a$ and $b$.
