I have two powers of binomials, $(1+x)^A$ and $(1+x)^B$.
By the binomial theorem theorem $(1+x)^A = \sum_{k}^{A}\binom{A}{k}x^k$ (1)
and $(1+x)^B = \sum_{k}^{B}\binom{B}{k}x^k$ (2)
Instead of this notation, for the summation (1) I write $\sum_{n}\binom{A}{n}x^n$
and for the summation (2) I write $\sum_{m}\binom{B}{m}x^m$.
Is the subindex $n$ the same than the index "$k=1$ to $A$" in (1) and the subindex $m$ the same than "$k=1$ to $B$" in (2)?
If I write $\sum_{n,m}\binom{A}{n}\binom{B}{m}x^{n+m}$ is the same than
$\sum_{k}^{A}\binom{A}{k}x^k \cdot \sum_{k}^{B}\binom{B}{k}x^k$?
If this change of notation is right, is it usual?
You can play fast & loose with your summation variables like that. But I find it helpful to keep tighter track of them ... \begin{eqnarray*} \left( \sum_{i=0}^{A} \binom {A}{i} x^i \right) \left( \sum_{j=0}^{B} \binom {B}{j} x^j \right) & =& (1+x)^A (1+x)^B =(1+x)^{A+B} \\ & =& \sum_{k=0}^{A+B} \binom {A+B}{k} x^k. \end{eqnarray*} So now calculating the product above and equating coefficients $x^k$ gives \begin{eqnarray*} \sum_{i+j=k} \binom {A}{i} \binom {B}{j} = \binom {A+B}{k}. \end{eqnarray*} Which is the Vandermonde convolution identity.