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.
 A: 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.

A: As you know, we usually derive the elements of each row as the sum of the two elements above the element in question. Therefore, you can work backwards by subtracting the upper left value from the lower value. When you apply this to the apex of the triangle and work upwards you do indeed get just  the answer shown above, which turns out to be a version of the original Pascal's Triangle, turned on its side, and with alternating signs.
