Continuing "Pascal's triangle" for negative binomial exponents

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.

Answer

There is a continuation respecting the addition law
$$\begin{align*} \binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1} \end{align*}$$

This way we can write

$$\begin{array}{l|rrrrrrrrrr} (1+x)^{-3}&\color{grey}{0}&\color{grey}{0}&1&-3x&6x^2&-10x^3&15x^4&-21x^5&38x^6&\ldots\\ (1+x)^{-2}&\color{grey}{0}&\color{grey}{0}&1&-2x&3x^2&-4x^3&5x^4&-6x^5&7x^6&\ldots\\ (1+x)^{-1}&\color{grey}{0}&\color{grey}{0}&1&-x&x^2&-x^3&x^4&-x^5&x^6&\ldots\\ (1+x)^{0}&\color{grey}{0}&\color{grey}{0}&1&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\ (1+x)^{1}&\color{grey}{0}&\color{grey}{0}&1&x&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\ (1+x)^{2}&\color{grey}{0}&\color{grey}{0}&1&2x&x^2&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\ (1+x)^{3}&\color{grey}{0}&\color{grey}{0}&1&3x&3x^2&x^3&\color{grey}{0}&\color{grey}{0}&\color{grey}{0}\\ \end{array}$$

For negative exponents $-n$ with $n>0$ we have
$$\begin{align*} (1+x)^{-n}&=\sum_{j=0}^\infty\binom{-n}{j}x^j =\sum_{j=0}^\infty\binom{n+j-1}{j}(-1)^jx^j\\ \end{align*}$$

Hint: See table 164 in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.