The Binomial Theorem

Expanding $(a+b)^{20}$ by multiplying out is impossible. The binomial theorem gives every coefficient directly — and, extended to fractional and negative powers, turns awkward expressions into series you can work with.

Pascal's triangle and combinations

The coefficients of $(a+b)^n$ are the entries in row $n$ of Pascal's triangle:

$$\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \end{array}$$

Each entry is the sum of the two above it. Algebraically the same numbers are the combinations:

$$\binom{n}{r} = {}^nC_r = \frac{n!}{r!(n-r)!}$$

$$\binom{5}{2} = \frac{5!}{2!\,3!} = \frac{120}{2 \times 6} = 10 \qquad \binom{8}{3} = \frac{8!}{3!\,5!} = 56$$

Two facts worth knowing: $\binom{n}{0} = \binom{n}{n} = 1$, and $\binom{n}{r} = \binom{n}{n-r}$ — the rows are symmetric.

The binomial expansion

For a positive integer $n$:

$$(a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$$

The powers of $a$ fall from $n$ to 0 while the powers of $b$ rise from 0 to $n$, and every term's powers add to $n$.

$$(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$$

$$(2+x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$$

since the terms are $\binom{5}{r}2^{5-r}x^r$: $32,\ 5(16),\ 10(8),\ 10(4),\ 5(2),\ 1$.

The general term

$$T_{r+1} = \binom{n}{r}a^{n-r}b^r$$

This is the tool for questions that ask for one coefficient out of a large expansion, without expanding anything else.

Finding a particular coefficient

Find the coefficient of $x^2$ in $(1+2x)^6$.

$$T = \binom{6}{2}(1)^4(2x)^2 = 15 \times 4x^2 = 60x^2$$

The coefficient is $60$.

Find the coefficient of $x^3$ in $(2+3x)^5$.

$$T = \binom{5}{3}(2)^2(3x)^3 = 10 \times 4 \times 27x^3 = 1080x^3$$

Find the term independent of $x$ in $\left(x^2 + \dfrac1x\right)^6$.

$$T_{r+1} = \binom{6}{r}\left(x^2\right)^{6-r}\left(\frac1x\right)^r = \binom{6}{r}x^{12-2r-r} = \binom{6}{r}x^{12-3r}$$

Independent of $x$ means the power is zero:

$$12 - 3r = 0 \quad \Rightarrow \quad r = 4$$

$$\binom{6}{4} = 15$$

The expansion for rational n

For any rational $n$, positive or not:

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$

valid only for $|x| < 1$.

When $n$ is a positive integer the series terminates — a factor of zero eventually appears — recovering the ordinary expansion. Otherwise it runs forever.

$(1+x)^{1/2}$:

$$1 + \frac12 x + \frac{\frac12\left(-\frac12\right)}{2}x^2 + \frac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{6}x^3 + \cdots = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots$$

$(1+x)^{-2}$:

$$1 - 2x + \frac{(-2)(-3)}{2}x^2 + \frac{(-2)(-3)(-4)}{6}x^3 + \cdots = 1 - 2x + 3x^2 - 4x^3 + \cdots$$

Approximations

A few terms of the series give an accurate value when $x$ is small.

Estimate $\sqrt{1.02}$.

Use $(1+x)^{1/2}$ with $x = 0.02$:

$$\sqrt{1.02} \approx 1 + \frac{0.02}{2} - \frac{(0.02)^2}{8} = 1 + 0.01 - 0.00005 = 1.00995$$

The true value is $1.0099504\ldots$, so three terms are accurate to seven decimal places.