## 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.

Pascal's triangle using 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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