Does there exist a pattern for the coefficients in a negative binomial expansion? This question is not about the explicit formula, but rather if there exist a continuation for Pascal's triangle.
$$\begin{array}l
(a+b)^{-2} &=&&&& \color{red}?\\
(a+b)^{-1} &=&&&& \color{red}?\\
(a+b)^{0} &=&&&& 1\\
(a+b)^{1} &=&&& 1a &+& 1b\\
(a+b)^{2} &=&& 1a^2 &+& 2ab &+&1b^2\\
(a+b)^{3} &=& 1a^3 &+& 3a^2b &+& 3ab^2 &+& 1b^3 &
\end{array}$$
It would obviously not be a triangle given that it's an infinite sum, but it seems reasonable that there should be a similar interpretation.
Best Answer
There is a continuation respecting the addition law \begin{align*} \binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1} \end{align*}
For negative exponents $-n$ with $n>0$ we have \begin{align*} (1+x)^{-n}&=\sum_{j=0}^\infty\binom{-n}{j}x^j =\sum_{j=0}^\infty\binom{n+j-1}{j}(-1)^jx^j\\ \end{align*}