# Intuitive explanation for negative binomial expansion

I understand the relationship between Pascal's triangle and binomial expansion for nonnegative integer powers by using combination e.g the binomial coefficient for $$x^3$$ in $$(a + x)^8$$ is equivalent to how many ways there are to choose 3 $$x$$'s from 8 possibilities.

I thought this idea would fail with regards to negative binomial expansion but I find out we can have cases where $$r<0$$ in $$nCr$$.

Can someone intuitively explain this? I kind of understand binomial expansion using Maclaurin series but I can't wrap my head around what negative combination is intuitively and why it works for negative binomial expansion.

Basically I don't understand the reason behind the combination formula still working for negative numbers.

• In the definition that I find most useful, $r$ is a non-negartive integer, but $n$ can take any real value. $\binom{n}{r}=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$. There are other definitions using the Ganma function for non-integer values of $r$, but they don't obey as many of the same relationships that the binomial coefficients with positive integral values do. – robjohn Feb 1 at 23:21
• The generalized binomial series naturally implies the properties of the generalized binomial coefficient. – William Feb 6 at 20:59
• You mean ${}_nC_r$ can have $n<0$, not $r<0$. – runway44 Mar 31 at 5:59
• @runway44: Yes. $n$ can take any real value, but $r$ is a non-negative integer. $_nC_r=\binom{n}{r}$ is the coefficient of $x^r$ in the Taylor series for $(1+x)^n$. – robjohn Apr 1 at 18:59

The binomial identity \begin{align*} \color{blue}{\binom{n+1}{k+1}=\binom{n}{k+1}+\binom{n}{k}}\tag{1} \end{align*} is not only valid for natural numbers $$n$$ but also for $$n\in\mathbb{Z}$$ and $$k\geq 0$$ a natural number. We can use this relationship to extend the Pascal triangle to negative numbers $$n$$ shown in the table below. The numbers of the Pascal triangle for $$n,k\geq 0$$ follow by setting $$\binom{n}{0}=1, \binom{0}{k}=\delta_{k,0}$$. The entries with negative $$n$$ follow by extending the rule $$\binom{n}{0}=1$$ for $$n\in\mathbb{Z}$$. \begin{align*} \begin{array}{r|rrrrrrrrl} k&0&1&2&3&4&5&6&&\\ \hline n&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&&\\ -4&1&\color{blue}{-4}&\color{blue}{10}&-20&35&-56&84&&(1+x)^{-4}\\ -3&1&-3&\color{blue}{6}&-10&15&-21&28&&(1+x)^{-3}\\ -2&1&-2&3&-4&5&-6&7&&(1+x)^{-2}\\ -1&1&-1&1&-1&1&-1&1&&(1+x)^{-1}\\ 0&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^0\\ 1&1&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^1\\ 2&1&2&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^2\\ 3&1&\color{blue}{3}&\color{blue}{3}&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^3\\ 4&1&4&\color{blue}{6}&4&1&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^4\\ \end{array} \end{align*}

When looking for instance at the last line of the table above we see the number $$6$$ is according to (1) the sum of the entry $$3$$ above $$6$$ with its left neighbor $$3$$. \begin{align*} 6=3+3\qquad\qquad \binom{4}{2}=\binom{3}{1}+\binom{3}{2} \end{align*} The algebraic connection is given by \begin{align*} \color{blue}{(1+x)^4=(1+x)(1+x)^3} \end{align*} or more detailed with $$[x^k]$$ denoting the coefficient of $$x^k$$: \begin{align*} \binom{4}{2}&=[x^2](1+x)^4\\ &=[x^2](1+x)(1+x)^3\\ &=\left([x^2]+[x^1]\right)(1+x)^3\\ &=\binom{3}{2}+\binom{3}{1}\tag{2} \end{align*} The algebraic connection holds also for negative $$n$$, for instance: \begin{align*} \color{blue}{(1+x)^{-3}=(1+x)(1+x)^{-4}} \end{align*} and we obtain similarly to (2): \begin{align*} 6=10+\left(-4\right)\qquad\qquad \binom{-3}{2}=\binom{-4}{2}+\binom{-4}{1} \end{align*} or more detailed with $$[x^k]$$ denoting the coefficient of $$x^k$$: \begin{align*} \binom{-3}{2}&=[x^2](1+x)^{-3}\\ &=[x^2](1+x)(1+x)^{-4}\\ &=\left([x^2]+[x^1]\right)(1+x)^{-4}\\ &=\binom{-4}{2}+\binom{-4}{1}\tag{2} \end{align*}

We can also see in the table the binomial identity \begin{align*} \binom{-n}{k}=\binom{n+k-1}{k}(-1)^k \end{align*} is given as rotation of a somewhat modified Pascal triangle. \begin{align*} \begin{array}{r|rrrrrrrrl} k&0&1&2&3&4&5&6&&\\ \hline n&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&&\\ -4&\color{blue}{1}&-4&10&-20&35&-56&84&&(1+x)^{-4}\\ -3&\color{blue}{1}&\color{blue}{-3}&6&-10&15&-21&28&&(1+x)^{-3}\\ -2&\color{blue}{1}&\color{blue}{-2}&\color{blue}{3}&-4&5&-6&7&&(1+x)^{-2}\\ -1&\color{blue}{1}&\color{blue}{-1}&\color{blue}{1}&\color{blue}{-1}&1&-1&1&&(1+x)^{-1}\\ 0&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^0\\ 1&1&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^1\\ 2&1&2&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^2\\ 3&1&3&3&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^3\\ 4&1&4&6&4&1&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^4\\ \end{array} \end{align*}

The Question

In this answer, it is shown that $$\binom{-n}{j}=(-1)^j\binom{n+j-1}{j}\tag1$$ If I understand correctly, the question is asking why, when $$(1)$$ is applied to the Binomial Theorem, we get $$(1+x)^{-n}$$.

Prior Results

In this answer are three proofs of this cancellation formula $$\sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k} =[n=k]\tag2$$ where $$[\cdots]$$ are Iverson Brackets.

Verification of $$\bf{(1)}$$ Using the Cauchy Product Formula

For $$k\ge1$$, we have \begin{align} \sum_{j=0}^k\binom{-n}{j}\binom{n}{k-j} &=\sum_{j=0}^k(-1)^j\binom{n+j-1}{n-1}\binom{n}{n+j-k}\tag{3a}\\ &=\sum_{j=0}^k(-1)^j\sum_{i=0}^{k-1}\binom{n+j-k}{i}\binom{k-1}{n-1-i}\binom{n}{n+j-k}\tag{3b}\\ &=\sum_{i=0}^{k-1}(-1)^{n-k-i}\binom{k-1}{n-1-i}[n=i]\tag{3c}\\[9pt] &=0\tag{3d} \end{align} Explanation:
$$\text{(3a)}$$: apply $$(1)$$ and the symmetry of Pascal's Triangle
$$\text{(3b)}$$: Vandermonde's Identity
$$\text{(3c)}$$: apply $$(2)$$
$$\text{(3d)}$$: $$\binom{k-1}{-1}=0$$

If $$k=0$$, the sum is $$\binom{-n}{0}\binom{n}{0}=1$$. Therefore, using $$(1)$$, $$(2)$$, and Vandermonde, we have shown $$\sum_{j=0}^k\binom{-n}{j}\binom{n}{k-j}=[k=0]\tag4$$ which by the Cauchy Product Formula verifies that $$\underbrace{\sum_{j=0}^\infty\binom{-n}{j}x^j}_{(1+x)^{-n}}\ \underbrace{\sum_{j=0}^n\binom{n}{j}x^j}_{(1+x)^n}=1\tag5$$

Verification of $$\bf{(1)}$$ Using Induction

Using the formula for the sum of a geometric series, we have $$(1+x)^{-1}=\sum_{k=0}^\infty(-1)^kx^k\tag6$$ We can verify $$(1)$$ inductively using $$(6)$$ and the Cauchy Product Formula. Assume that $$(1)$$ is true for a given $$n$$, then \begin{align} (1+x)^{-n-1} &=\color{#C00}{(1+x)^{-1}}\color{#090}{(1+x)^{-n}}\tag{7a}\\[9pt] &=\sum_{k=0}^\infty\sum_{j=0}^k\color{#C00}{(-1)^{k-j}x^{k-j}}\color{#090}{(-1)^j\binom{n+j-1}{j}x^j}\tag{7b}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^k\binom{n+j-1}{j}x^k\tag{7c}\\ &=\sum_{k=0}^\infty(-1)^k\binom{n+k}{k}x^k\tag{7d}\\ \end{align} Explanation:
$$\text{(7b)}$$: $$(6)$$ and the inductive hypothesis
$$\text{(7c)}$$: collect terms
$$\text{(7d)}$$: Hockey-Stick Identity

which verifies $$(1)$$ for $$n+1$$.

$$(1+x)^{-n}$$ can also be written as $$\frac{1}{(1+x)^n}$$.

The denominator can be expanded and then viewed as the usual identity $$\frac1{1-x}=1+x+x^2+\dots$$

For example with $$n=2$$:

$$1+(-2x-x^2)+(2x+x^2)^2+(-2x-x^2)^3+\dots$$

This can have the like terms collected, which gives:

$$\binom{-2}{1}=-2$$ $$\binom{-2}{2}=3$$

etc...