## Content

### The binomial theorem

We are now ready to prove the binomial theorem. We will give another proof later in the module using mathematical induction.

###### Theorem (Binomial theorem)

For each positive integer \(n\),

\[ (a+b)^n = a^n + \dbinom{n}{1}a^{n-1}b + \dbinom{n}{2}a^{n-2}b^2+\dots+\dbinom{n}{r}a^{n-r}b^r+\dots+ \dbinom{n}{n-1}ab^{n-1} + b^n. \]###### Proof

Suppose that we have \(n\) factors each of which is \(a+b\). If we choose one letter from each of the factors of

\[ (a+b)(a+b)(a+b)\dotsm(a+b) \]and multiply them all together, we obtain a term of the product. If we do this in every possible way, we will obtain all of the terms.

- If we choose \(a\) from every one of the factors, we get \(a^n\). This can only be done in one way.
- We could choose \(b\) from one of the factors and choose \(a\) from the remaining \(n-1\) factors. The number of ways of choosing one \(b\) from \(n\) factors is \(\smash{\dbinom{n}{1}}\). So the term with \(b\) is \(\dbinom{n}{1}a^{n-1}b\).
- We could choose \(b\) from two of the factors and choose \(a\) from the remaining \(n-2\) factors. The number of ways of choosing two \(b\)'s from \(n\) factors is \(\smash{\dbinom{n}{2}}\). So the term with \(b^2\) is \(\dbinom{n}{2}a^{n-2}b^2\).
- In general, we choose \(b\) from \(r\) factors and choose \(a\) from the remaining \(n-r\) factors. The number of ways of choosing \(r\) \(b\)'s from \(n\) factors is \(\smash{\dbinom{n}{r}}\). So the term with \(b^r\) is \(\dbinom{n}{r}a^{n-r}b^r\).
- If we choose \(b\) from every one of the factors, we get \(b^n\). This can be done in only one way.

Thus,

\((a+b)^n = a^n + \dbinom{n}{1}a^{n-1}b + \dbinom{n}{2}a^{n-2}b^2+\dots+\dbinom{n}{r}a^{n-r}b^r+\dots+ \dbinom{n}{n-1}ab^{n-1} + b^n\).\(\Box\)

The binomial theorem can also be stated using summation notation:

\[ (a+b)^n = \sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r. \]Substituting with \(a=1\) and \(b=x\) gives

\[ (1+x)^n = \dbinom{n}{0}+ \dbinom{n}{1}x + \dbinom{n}{2}x^2+\dots+\dbinom{n}{r}x^r+\dots+ \dbinom{n}{n-1}x^{n-1} + \dbinom{n}{n}x^n. \]We can now display Pascal's triangle with the notation of the binomial theorem.

\(n\) | \(x^0\) | \(x^1\) | \(x^2\) | \(x^3\) | \(x^4\) | \(x^5\) | \(x^6\) | \(x^7\) | \(x^8\) |
---|---|---|---|---|---|---|---|---|---|

0 | \(\dbinom{0}{0}\) | ||||||||

1 | \(\dbinom{1}{0}\) | \(\dbinom{1}{1}\) | |||||||

2 | \(\dbinom{2}{0}\) | \(\dbinom{2}{1}\) | \(\dbinom{2}{2}\) | ||||||

3 | \(\dbinom{3}{0}\) | \(\dbinom{3}{1}\) | \(\dbinom{3}{2}\) | \(\dbinom{3}{3}\) | |||||

4 | \(\dbinom{4}{0}\) | \(\dbinom{4}{1}\) | \(\dbinom{4}{2}\) | \(\dbinom{4}{3}\) | \(\dbinom{4}{4}\) | ||||

5 | \(\dbinom{5}{0}\) | \(\dbinom{5}{1}\) | \(\dbinom{5}{2}\) | \(\dbinom{5}{3}\) | \(\dbinom{5}{4}\) | \(\dbinom{5}{5}\) | |||

6 | \(\dbinom{6}{0}\) | \(\dbinom{6}{1}\) | \(\dbinom{6}{2}\) | \(\dbinom{6}{3}\) | \(\dbinom{6}{4}\) | \(\dbinom{6}{5}\) | \(\dbinom{6}{6}\) | ||

7 | \(\dbinom{7}{0}\) | \(\dbinom{7}{1}\) | \(\dbinom{7}{2}\) | \(\dbinom{7}{3}\) | \(\dbinom{7}{4}\) | \(\dbinom{7}{5}\) | \(\dbinom{7}{6}\) | \(\dbinom{7}{7}\) | |

8 | \(\dbinom{8}{0}\) | \(\dbinom{8}{1}\) | \(\dbinom{8}{2}\) | \(\dbinom{8}{3}\) | \(\dbinom{8}{4}\) | \(\dbinom{8}{5}\) | \(\dbinom{8}{6}\) | \(\dbinom{8}{7}\) | \(\dbinom{8}{8}\) |

The \(n\)th row of this table gives the coefficients of \((1+x)^n\), where \(\dbinom{n}{r}\) is the coefficient of \(x^r\) in this expansion. Numbers of the form \(\smash{\dbinom{n}{r}}\) are called **binomial coefficients**.