Expand expressions of the form a plus b to the power n for a positive integer n using the binomial theorem and binomial coefficients

What this dot point is asking

SEAB wants you to expand $(a + b)^n$ for a positive whole number $n$ using the binomial theorem, rather than multiplying the bracket out repeatedly. You need the binomial coefficients (from Pascal's triangle or the $\binom{n}{r}$ notation) and the pattern of decreasing powers of $a$ and increasing powers of $b$.

The answer

The structure of the expansion

The expansion of $(a + b)^n$ has $n + 1$ terms. In each term the powers of $a$ and $b$ add to $n$: $a$ starts at power $n$ and falls to $0$, while $b$ rises from $0$ to $n$:

The binomial coefficients

The coefficient $\binom{n}{r}$ (read "n choose r") counts the terms and is given by:

For small $n$ they are read from Pascal's triangle, where each entry is the sum of the two above it. Row $4$ is $1, 4, 6, 4, 1$.

Handling signs and coefficients

When $b$ is itself a product such as $-2x$, substitute the whole thing and apply the powers carefully: $(-2x)^2 = 4x^2$ (positive), $(-2x)^3 = -8x^3$ (negative). Sign errors here are the most common mistake.

Ascending or descending powers

Expand in ascending powers of $x$ (smallest power first) or descending, as the question asks. "Up to the term in $x^2$" means write only the first three terms.

The symmetry of the coefficients

Each row of Pascal's triangle is symmetric: $\binom{n}{r} = \binom{n}{n-r}$, so the coefficients read the same forwards and backwards. This is a useful check, since a row that is not symmetric has an error, and it lets you write the second half of a long row from the first.

Why the powers add to n

Every term picks either $a$ or $b$ from each of the $n$ brackets in $(a + b)^n$, so the powers of $a$ and $b$ in a term must total $n$. The coefficient $\binom{n}{r}$ counts how many ways to choose the $r$ brackets that contribute a $b$, which is why the same numbers appear in counting problems.

Examples in context

Example 1. Approximating a power. Expanding $(1 + x)^n$ for small $x$ and keeping only the first few terms gives a quick approximation, for instance $(1.02)^5 \approx 1 + 5(0.02) = 1.10$, the idea behind small-change estimates.

Example 2. Probability of repeated trials. The binomial coefficients $\binom{n}{r}$ count the ways an event can occur $r$ times in $n$ trials, linking the algebraic expansion to the binomial probability you meet in statistics.

Try this

Q1. Write down the binomial coefficients for $(a + b)^5$. [1 mark]

• Cue. Row $5$ of Pascal's triangle: $1, 5, 10, 10, 5, 1$.

Q2. Expand $(1 + x)^3$. [2 marks]

• Cue. $1 + 3x + 3x^2 + x^3$.

Q3. Find the first three terms of $(2 - x)^4$ in ascending powers of $x$. [3 marks]

• Cue. $16 + 4(8)(-x) + 6(4)x^2 = 16 - 32x + 24x^2$.