General Binomial Expansion

Question

I have two operators $A$ and $B$ that do not commute, thus the binomial expansion can't be used. Is there any general formula for $(A+B)^n$, for $n \in \mathbb{N}$?

Thank you.

Answer 1

As Will wrote in his since-deleted answer. it's the sum of all the $2^n$ words of length $n$ over the alphabet $A,B$. You won't get a simpler answer, unless you know something useful about $A$ and $B$.

Answer 2

I am not sure if you will find this useful, but I think the best you can do is: $$ (A+B)^n = \sum_{f:\{1,\dotsc, n\}\to \{A, B\}} f(1)f(2)\dotsb f(n) $$ The sum is over all functions from the set $\{1,\dotsc,n\}$ to the set $\{A, B\}$.

When $A$ and $B$ commute, the coefficient of $A^kB^{n-k}$ is the number of such functions which take the value $A$ $k$ times and take the value $B$ $n-k$ times, giving the binomial theorem.

Answer 3

If the variables $A$ and $B$ are free (they have no relations), then the expansion of $(A + B)^n$ has $2^n$ terms. They are all $n$-letter words in the alphabet $\{A, B\}$.

Answer 4

Thanks everyone for replying. In conclusion, the answer to my question would be a no. Unless of course, there is some other info on $A$ and $B$.