Finding one term of a binomial expansion

Exam questions rarely want all nine terms of $(2x - 1)^8$. They want one: "the coefficient of $x^3$", "the constant term". So go straight to it.

The term containing $b^r$ is

$\binom{n}{r}\,a^{\,n-r}\,b^{\,r}$

Two powers, one coefficient, and the powers add up to $n$. Pick $r$ to get the power of $x$ you want.

What the symbols mean:

• $n$ is the power the whole bracket is raised to.
• $a$ is the first thing in the bracket and $b$ is the second, including its sign. In $(2 - 3x)^6$, $a = 2$ and $b = -3x$.
• $r$ is how many times $b$ is used in this term ($r = 0, 1, 2, \dots, n$).
• $\binom{n}{r}$, read "$n$ choose $r$" (also written ${}^nC_r$), is the number of ways to choose $r$ things from $n$. It equals $\dfrac{n!}{r!\,(n - r)!}$, where $n! = n \times (n - 1) \times \dots \times 2 \times 1$. Your calculator has an $nCr$ button: $\binom{6}{3} = \dfrac{6!}{3!\,3!} = \dfrac{720}{36} = 20$.

Why this formula? $(a + b)^n$ means $n$ brackets multiplied together. To make one term you pick either $a$ or $b$ from every bracket. If you pick $b$ from $r$ of them, you pick $a$ from the other $n - r$, giving $a^{n-r}b^r$. And there are $\binom{n}{r}$ ways to decide which $r$ brackets supply the $b$, so that many copies of $a^{n-r}b^r$ add up.