Introduction

From Squares and Cubes to Higher Powers

In earlier classes you learned to expand squares and cubes of binomials such as $(a+b)^2$, $(a-b)^2$, $(a+b)^3$, and $(a-b)^3$. These ready-made expansions double as a numerical shortcut: if you can write a number as a sum or difference of two convenient numbers, you can find its square or cube without a long multiplication. For instance, $(98)^2$ is really $(100-2)^2$, and $(999)^3$ is really $(1000-1)^3$ — expanding the binomial gives the value directly, without ever multiplying $98 \times 98$ or $999 \times 999 \times 999$ by hand.

Where This Shortcut Breaks Down

The trouble starts when the exponent itself grows. Try the same idea on $(98)^5$ or $(101)^6$: there is no ready-made "square" or "cube" identity to fall back on, and expanding the bracket by repeated multiplication — term by term, five or six times over — quickly becomes long and easy to get wrong.

The Binomial Theorem: One Rule for Any Power

This is exactly the gap the binomial theorem closes. It gives a single, systematic rule for expanding $(a+b)^n$ for any power $n$, without ever multiplying the bracket out by hand. In its most general form, the theorem even holds when $n$ is a rational number.

What's Ahead in This Chapter

Starting from the familiar expansions of $(a+b)^0$ through $(a+b)^4$, the sections that follow build up the full pattern step by step — first using Pascal's triangle, then a compact formula written with combinations — until you have a single rule that expands $(a+b)^n$ for any positive integer $n$, along with a formal proof and a set of useful special cases.