# 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.

• For that, you should first explain what you would put there. Are you aiming to write $(a+b)^{-n} = \sum_{i,j = 0}^{\infty} p_{ij}a^ib^j$ and want to put the $p_{ij}$ somewhere in there? Can you show that such an expansion always exist with real (or even integer) coefficients? How would you arrange these numbers in a single line, which ordering would you take. So there are many questions to specify before considering how it would look up there. – Dirk May 24 '17 at 10:41
• I don't think it is possible to avoid infinite series. – AHB May 24 '17 at 10:44
• There are negative binomial coefficients such that we have the expansion $(a+b)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} a^{k}b^{-k-n}$. Those numbers also satisfy Pascal's identity $\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$. – Joel Cohen May 24 '17 at 10:46
• There is a well known formula for the entries which uses the factorial function. The factorial function can be generalised with the gamma function so you could generalise the binomial exponents to complex numbers. – badjohn May 24 '17 at 10:57
• @AHB I didn't exclude infinite series anywhere. Pascal's triangle is an infinite series itself along the diagonals. – Frank Vel May 24 '17 at 10:57

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.

• So for $(y+x)^n$, the exponent of $y$ depends on which diagonal it lies in? – Frank Vel May 24 '17 at 18:57
• @FrankVel: Not quite clear to me which diagonal you address. Consider factoring out $y^{-n}$ From $(x+y)^{-n}$ which gives $y^{-n}(1+x/y)^{-n}$. Expanding, looking at a term $\binom{-n}{j}(x/y)^j$ and multiplication with $y^{-n}$ gives $\binom{-n}{j}x^jy^{-n-j}$. – Markus Scheuer May 24 '17 at 19:18
• Well for the top-left->bottom-right diagonal, the $y^k$ diagonal will be the $k$th diagonal below the $... 0 0 0 1 0 0 0 ...$ diagonal. If you mirror this on the $.. 0 0 1 0 0 0...$ row, you will get the same pattern above for the bottom-left->top-right diagonals? – Frank Vel May 24 '17 at 19:46
• @FrankVel: Now I see what you mean and you're right. Respecting the sign, the symmetry is given. – Markus Scheuer May 24 '17 at 19:57

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.