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%TCIDATA{<META NAME="Title" CONTENT="Reference/Mathematics/The Binomial Theorem">}
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\section{The Binomial Theorem\label{The Binomial Theorem}}

An expression of the form $a+b$ is called a \textbf{binomial}. Although in
principle it is easy to raise $a+b$ to any power, raising it to a very high
power would be a tedious task. There is a formula that gives the expansion
of $\left( a+b\right) ^{n}$ for any natural number $n.\ $\ To discover this
formula, find a pattern for the successive powers of $a+b$.

\begin{center}
\begin{tabular}{lllllll}
$\left( a+b\right) ^{1}$ & $=$ & $a+b$ &  &  &  &  \\ 
$\left( a+b\right) ^{2}$ & $=$ & $a^{2}+2ab$ & $+b^{2}$ &  &  &  \\ 
$\left( a+b\right) ^{3}$ & $=$ & $a^{3}+3a^{2}b$ & $+3ab^{2}$ & $+b^{3}$ & 
&  \\ 
$\left( a+b\right) ^{4}$ & $=$ & $a^{4}+4a^{3}b$ & $+6a^{2}b^{2}$ & $+4ab^{3}
$ & $+b^{4}$ &  \\ 
$\left( a+b\right) ^{5}$ & $=$ & $a^{5}+5a^{4}b$ & $+10a^{3}b^{2}$ & $%
+10a^{2}b^{3}$ & $+5ab^{4}$ & $+b^{5}$ \\ 
& $\vdots $ &  &  &  &  & 
\end{tabular}
\end{center}

The following simple patterns emerge for the expansion of $\left( a+b\right)
^{n}$:

\begin{enumerate}
\item  There are $n+1$ terms, the first being $a^{n}$ and the last being $%
b^{n}$.

\item  The exponents of $a$ decrease by $1$ from term to term while the
exponents of $b$ increase by $1$.

\textbf{The exponents of }$a$\textbf{\ decrease:} 
\[
\left( a+b\right)
^{5}=a^{5}+5a^{4}b^{1}+10a^{3}b^{2}+10a^{2}b^{3}+5a^{1}b^{4}+b^{5}
\]

\textbf{The exponents of }$b$\textbf{\ increase:} 
\[
\left( a+b\right)
^{5}=a^{5}+5a^{4}b^{1}+10a^{3}b^{2}+10a^{2}b^{3}+5a^{1}b^{4}+b^{5}
\]

\item  The sum of the exponents of $a$ and $b$ in each term is $n$.
\end{enumerate}

With these observations you can write the form of the expansion of $\left(
a+b\right) ^{n}$ for any natural number $n$. For example, writing a question
mark for the missing coefficients you know 
\[
\left( a+b\right)
^{8}=a^{8}+?a^{7}b+?a^{6}b^{2}+?a^{5}b^{3}+?a^{4}b^{4}+?a^{3}b^{5}+?a^{2}b^{6}+?ab^{7}+b^{8} 
\]

To complete the expansion you need to determine these coefficients. To find
a pattern, write the coefficients in the expansion of $(a+b)^{n}$ for the
first few values of $n$ in a triangular array as shown in the following
array, which is called \textbf{Pascal's triangle\label{Pascal's triangle}.} 
\[
\begin{array}{lllllllllllll}
\left( a+b\right) ^{0} &  &  &  &  &  &  & 1 &  &  &  &  &  \\ 
\left( a+b\right) ^{1} &  &  &  &  &  & 1 &  & 1 &  &  &  &  \\ 
\left( a+b\right) ^{2} &  &  &  &  & 1 &  & 2 &  & 1 &  &  &  \\ 
\left( a+b\right) ^{3} &  &  &  & 1 &  & 3 &  & 3 &  & 1 &  &  \\ 
\left( a+b\right) ^{4} &  &  & 1 &  & 4 &  & 6 &  & 4 &  & 1 &  \\ 
\left( a+b\right) ^{5} &  & 1 &  & 5 &  & 10 &  & 10 &  & 5 &  & 1
\end{array}
\]
The row corresponding to $(a+b)^{0}$ is called the zeroth row and is
included to show the symmetry of the array.

\subsection{Key Property of Pascal's Triangle}

\begin{itemize}
\item  Every entry (other than a $1$) is the sum of the two entries
diagonally above it.
\end{itemize}

Pascal's triangle is not practical for finding $(a+b)^{n}$ for large values
of $n$. The reason is that the method used for finding the successive rows
of Pascal's triangle is recursive. Thus, to find the 100th row of this
triangle we must first find all the preceding rows.

Examining the pattern in the coefficients more carefully shows a formula
that calculates directly any coefficient in the binomial expansion.

\subsection{The Binomial Coefficient\label%
{Definition of binomial coefficient}}

Let $n$ and $r$ be nonnegative integers with $r\leq n$. The \textbf{binomial
coefficient} is denoted by $\binom{n}{r}$ and is defined by 
\[
\binom{n}{r}=\frac{n!}{r!\left( n-r\right) !} 
\]
\CustomNote{Note}{%
Note that this is the same formula as that for \hyperref{combinations}{}{}{%
Al08_03.tex#Combinations of n objects taken r at a time}. In fact, if you
sit down and examine carefully where the $a^{r}b^{n-r}$ terms in the
expansion of $\left( a+b\right) ^{n}$ come from, this offers another way to
see why the Binomial Theorem is true. For you're multiplying 
\[
\stackunder{n\text{ factors}}{\underbrace{\left( a+b\right) \left(
a+b\right) ...\left( a+b\right) }} 
\]
and by the distributive law this boils down to taking all possible products
of one or the other of $a$ and $b$ from each factor (like $a$ from the first
factor, $b$ from the second, $b$ from the third, etc.) and adding them
together. Now each $a^{r}b^{n-r}$ term comes from a combination involving
exactly $r$ choices of $a$ (the rest, of course, being $b$'s). So how many
such terms are there? How many ways can you select exactly $r$ factors to
use the $a$ terms from? Why, of course, it's $_{n}C_{r}$, the number of ways
to choose $r$ objects from $n$.}

\subsection{Key Property of the Binomial Coefficients}

For any nonnegative integers $r$ and $k$ with $r\leq k$, 
\[
\binom{k}{r-1}+\binom{k}{r}=\binom{k+1}{r} 
\]

\subsection{The Binomial Theorem\label{Statement of the Binomial Theorem}}

\[
(a+b)^{n}=\binom{n}{0}a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}%
a^{n-2}b^{2}+\cdot \cdot \cdot +\binom{n}{n-1}ab^{n-1}+\binom{n}{n}b^{n}
\]
So now you can fill in the question marks in the expansion of $\left(
a+b\right) ^{8}$: 
\begin{eqnarray*}
\left( a+b\right) ^{8} &=&\binom{8}{0}a^{8}+\binom{8}{1}a^{7}b+\binom{8}{2}%
a^{6}b^{2}+\binom{8}{3}a^{5}b^{3}+\binom{8}{4}a^{4}b^{4}+\binom{8}{5}%
a^{3}b^{5}+\binom{8}{6}a^{2}b^{6}+\binom{8}{7}ab^{7}+\binom{8}{8}b^{8} \\
&=&a^{8}+8a^{7}b+28a^{6}b^{2}+56a^{5}b^{3}+70a^{4}b^{4}+56\allowbreak
a^{3}b^{5}+28a^{2}b^{6}+8ab^{7}+b^{8}
\end{eqnarray*}

\subsubsection{General Term of the Binomial Expansion}

The term that contains $a^{r}$ in the expansion of $(a+b)^{n}$ is 
\[
\binom{n}{n-r}a^{r}b^{n-r}
\]

\subsection{Related Scientific Notebook Topics}

{\small \hyperref{Binomials}{}{}{../../Help/Binomials.tex}}

{\small \hyperref{Factorial}{}{}{../../Help/DM2-2.tex#Factorial}}

\noindent \hrulefill

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\noindent \hrulefill

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