Binomial Expansion for non-commutative setting

I don't know if you prefer a particular presentation of the formula, but this is essentially covered by the Baker-Campbell-Hausdorff formula, or actually it's dual, Zassenhaus formula, which in your case reduces to $$e^{(A+B)t}=e^{At}e^{Bt}e^{-[A,B]t^2/2},$$ where one side is the generating function for $(A+B)^n$ while the other has terms of the form $f(n,m,p)A^nB^mC^p$. The binomial theorem here is given by equating the coefficients of $t^n$ on both sides. $$(A+B)^n=\sum_{n\equiv k\pmod{2}} \left(\sum_{r=0}^k \binom{k}{r}A^rB^{k-r}\right)\left(-\frac{C}{2}\right)^{\frac{n-k}{2}}\frac{n!}{k!(\frac{n-k}{2})!}$$

Here are the calculations steps, in case it can help anyone:

Left hand side can be written $$e^{t(a+b)} = \sum_{k} \frac{(a+b)^k t^k}{k!}$$ where one recognizes the term we want to compute.

Right hand side is given by $$e^{ta} e^{tb} e^{- c t^2 / 2} = \sum_{i, j, k} t^{i + j + 2k} \frac{a^i b^j (-c / 2)^k}{i! \, j! \, k!}$$

Identifying both sides, one has: $$\frac{(a+b)^n}{n!} = \sum_{i + j + 2k = n} \frac{(-c / 2)^k}{i! \, j! \, k!} a^i b^j$$

$$\frac{(a+b)^n}{n!} = \sum_{\substack{i + j \leq n \\ i+j = n [2]}} \frac{(-c / 2)^{(n - i - j) / 2}}{i! \, j! \, \left(\frac{n - i - j}{2}\right)!} a^i b^j$$

Let us note $k = m+n$ and $r = m$, $$\frac{(a+b)^n}{n!} = \sum_{\substack{0 \leq r \leq k \leq n \\ k = n [2]}} \frac{(-c / 2)^{(n - k) / 2}}{\left(\frac{n - k}{2}\right)! \, r! \, (k-r)!} a^r b^{k-r}$$

$$\frac{(a+b)^n}{n!} = \sum_{\substack{k = 0 \\ k = n [2]}}^n \frac{(-c / 2)^{(n - k) / 2}}{k! \left(\frac{n - k}{2}\right)!} \sum_{r = 0}^{k} \binom{k}{r} a^r b^{k-r}$$

Which is the expected formula.