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.
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