the binomial theorem

The binomial theorem gives the values of the coefficients of the expansion of:

    \begin{align*} (a+b)^n&=(a+b)(a+b)(a+b)\dots(a+b)\\ \end{align*}

where n is any positive integer. (Newton gave the formula for any rational n.)
The picture above shows the expansions for n=2 and 3.
The coefficients are the same as the rows of Pascal’s triangle.

\displaystyle(a+b)(a+b)(a+b)\dots(a+b) is multiplied out by choosing one number from each bracket.
There is one way of choosing a^n — choosing a in each bracket, similarly one way of choosing b^n.
There are n ways of choosing a^{n-1}b — one for each bracket the b is chosen from.
These are combinations of n elements, (in the strict mathematical sense of “combination”,) in which \alpha identical elements a and \beta identical elements b occur, where \alpha+\beta=n.
The number of them is given by \displaystyle C=\frac{n!}{\alpha!\beta!}.
(0! is defined as 1, so that the formula gives the right answer for the number of occurrences of \displaystyle a^n=\frac{n!}{n!0!}=1 and \displaystyle b^n=\frac{n!}{0!n!}=1.)

So now we have the binomial expansion:

    \[(a+b)^n=\sum_{\alpha+\beta=n} \frac{n!}{\alpha!\beta!} a^\alpha b^\beta\]

For example,

    \begin{align*} (a+b)^5&=a^5+\frac{5!}{4!1!}a^4b+\frac{5!}{3!2!}a^3b^2+\frac{5!}{2!3!}a^2b^3+\frac{5!}{1!4!}ab^4+b^5\\[8pt] &=a^5+\frac{5\times4}{4}a^4b+\frac{5\times4}{2}a^3b^2+\frac{5\times4}{2}a^2b^3+\frac{5\times4}{4}ab^4+b^5\\[8pt] &=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\\ \end{align*}

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\displaystyle\frac{n!}{\alpha!\beta!}=\frac{n(n-1)(n-2)\dots(n-\alpha+1)}{1\cdot 2\cdot 3\dots \alpha}\quad is usually written \displaystyle\quad\binom{n}{\alpha}


Uh, or there’s Peter’s Method.

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