In some questions about polynomials, I've found the following formula:
$\binom{a}{k}=\sum_{j=0}^k\binom{a-b}{j}\binom{b}{k-j}$
This can be gotten of the coeficient of $x^k$ in $(1+x)^a$ and in $(1+x)^{a-b}(1+x)^b$... Right?
My uncertainty here is about the cases with $a-b<k$ or $b<k$. So I'll might define $\binom{n}{m}$ with $m>n$. Is it zero? I've done some calculus and I think it's zero to maintain the formula true.
Other example:
In one question, the polynomial was wroted by
$f(x)=A_0+A_1\binom{x}{1}+A_2\binom{x}{2}+\cdots+A_n\binom{x}{n}$
So, how I can write, for instance, $f(1)$? It might be
$f(1)=A_0+A_1\binom{1}{1}+A_2\binom{1}{2}+\cdots+A_n\binom{1}{n}$
and all the terms since the second will be zero?
So $f(1)=A_0+A_1$?
This difficulty aroused a curiosity if the possibility of binomials were extended to negative integers as well. Is there, perhaps?
Thank you very much.