Binomial Theorem for Negative Index or Fraction

Why a new form is needed. The Binomial Theorem of Section 4.2 was stated using $^nC_r=\dfrac{n!}{r!(n-r)!}$, which requires $n$ to be a non-negative integer so that $n!$ is defined. When $n$ is negative or a fraction, $n!$ has no meaning, so the theorem must be restated without factorials of $n$: $(a+b)^n=a^n+\dfrac{n}{1!}a^{n-1}b+\dfrac{n(n-1)}{2!}a^{n-2}b^2+\dfrac{n(n-1)(n-2)}{3!}a^{n-3}b^3+\cdots+\dfrac{n(n-1)\cdots(n-r+1)}{r!}a^{n-r}b^r+\cdots+b^n,$ so that the general term is $t_{r+1}=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}a^{n-r}b^r$ — built from a product of $r$ consecutive factors starting at $n$ and counting down, divided by $r!$, which makes sense for any real $n$, not just a positive integer.

Specialising to $a=1,b=x$ recovers the familiar finite form when $n\in N$: $(1+x)^n=1+{}^nC_1x+{}^nC_2x^2+\cdots+x^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots+x^n$, a finite sum since the product $n(n-1)\cdots(n-r+1)$ eventually contains the factor $0$ once $r>n$.

The extension to negative or fractional $n$ (stated without proof). For $|x|<1$, $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots+\dfrac{n(n-1)\cdots(n-r+1)}{r!}x^r+\cdots,$ where now $n$ need not be an integer and the series on the right does not terminate — it has infinitely many terms, and the restriction $|x|<1$ is exactly what makes the infinite sum converge to a finite value. The general term here is $t_{r+1}=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}x^r$, $r\ge0$.

Remarks. (1) Replacing $x$ by $-x$: for $|x|<1$ and any real $n$ that is not a positive integer, $(1-x)^n=1-nx+\dfrac{n(n-1)}{2!}x^2-\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots$, with general term $t_{r+1}=(-1)^r\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}x^r$. (2) For a general $(a+b)^n$ with real $n$ and $|b|<|a|$: factor out the larger term, $(a+b)^n=a^n\left(1+\dfrac ba\right)^n$, and expand the bracket by the series above with $x=b/a$ (note $|b/a|<1$ exactly because $|b|<|a|$, so the series converges). Note. When expanding $(a+b)^n$ for a negative or fractional $n$, always first reduce it to the standard form where the leading term inside the bracket is $1$ and the second term is numerically smaller than $1$ — this is precisely what taking out $a^n$ and forming $b/a$ achieves.

Six particular expansions to recognise, all for $|x|<1$:

1. $\dfrac{1}{1+x}=(1+x)^{-1}=1-x+x^2-x^3+x^4-x^5+\cdots$
2. $\dfrac{1}{1-x}=(1-x)^{-1}=1+x+x^2+x^3+x^4+x^5+\cdots$
3. $\dfrac{1}{(1+x)^2}=(1+x)^{-2}=1-2x+3x^2-4x^3+\cdots$
4. $\dfrac{1}{(1-x)^2}=(1-x)^{-2}=1+2x+3x^2+4x^3+\cdots$
5. $(1+x)^{1/2}=1+\dfrac12x-\dfrac18x^2+\dfrac{1}{16}x^3-\cdots$
6. $(1-x)^{1/2}=1-\dfrac12x-\dfrac18x^2-\dfrac{1}{16}x^3-\cdots$

Solved Example 1. State the first four terms in the expansion of $(a-b)^{-4}$, where $|b|<|a|$. Solution. $(a-b)^{-4}=a^{-4}\left(1-\dfrac ba\right)^{-4}=a^{-4}\left[1+(-4)\left(-\dfrac ba\right)+\dfrac{(-4)(-5)}{2!}\left(-\dfrac ba\right)^2+\dfrac{(-4)(-5)(-6)}{3!}\left(-\dfrac ba\right)^3+\cdots\right]=\dfrac{1}{a^4}+\dfrac{4b}{a^5}+\dfrac{10b^2}{a^6}+\dfrac{20b^3}{a^7}+\cdots$.

Solved Example 2. State the first four terms in the expansion of $(a+b)^{-1}$, $|b|<|a|$. Solution. $(a+b)^{-1}=a^{-1}\left(1+\dfrac ba\right)^{-1}=a^{-1}\left[1-\dfrac ba+\dfrac ba^2-\cdots\right]=\dfrac1a-\dfrac{b}{a^2}+\dfrac{b^2}{a^3}-\dfrac{b^3}{a^4}+\cdots$. …