Binomial Formula (C)

We will have on many occasions the need to expand expressions of the form

(a + b)^N

where N is a positive integer. There is a simple formula for this, known as the binomial formula, which says that

Theorem 2864 (Binomial Formula)

If N is a positive integer and a and b are complex numbers then

$$(a+b)^N = \sum_{k=0}^N\mbox{$\left(\begin{array} {c}N\ k\end{array}\right)$}a^kb^{N-k}$$

where

$$\mbox{$\left(\begin{array} {c}N\ k\end{array}\right)$} = \frac{N!}{k!(N-k)!}.$$

It is not surprising that the terms in (a + b)^N involve powers of a and b. The remarkable thing is the coefficients. The theorem becomes transparent when one realizes that $\mbox{$\left(\begin{array} {c}N\ k\end{array}\right)$}$ counts the number of ways of choosing k things out of N things. Thus, to obtain a^kb^N-k one had to choose k a's out of the N factors in (a+b)^N, and there are $\mbox{$\left(\begin{array} {c}N\ k\end{array}\right)$}$ ways to do this.

Here is an example:

$$(a+b)^4 = \mbox{$\left(\begin{array} {c}4\ 0\end{array}\rig... ... +\mbox{$\left(\begin{array} {c}4\ 4\end{array}\right)$}a^4b^0$$

so

(a+b)⁴ = b⁴ + 4ab³ + 6a²b² + 4a³b +a⁴.

Remark: The binomial coefficients

$$\mbox{$\left(\begin{array} {c}p\ k\end{array}\right)$}$$

can be given a meaning even if p is not a positive integer, so long as k is positive integer.

We put

$$\mbox{$\left(\begin{array} {c}p\ 0\end{array}\right)$} = 1$$

and

$$\mbox{$\left(\begin{array} {c}p\ 1\end{array}\right)$} = p$$

For $${k \geq 2}$$

$$\mbox{$\left(\begin{array} {c}p\ k\end{array}\right)$} = \frac{p(p-1)\cdots(p-(k-1))}{k!}.$$

This conforms to the case where p is a positive integer since

$$p! = p(p-1)\cdots(p-(k-1))(p-k)!$$

For example

$$\mbox{$\left(\begin{array} {c}\frac{1}{3}\ 2\end{array}\right)$} = \frac{\frac{1}{3}\frac{-2}{3}}{2!}$$