EN1 - The binomial expansion

A power such as $(2-3x)^5$ looks like a long multiplication, but counting tells you its terms directly. You will see why combinations appear in an expansion, expand a positive integer power, and find a requested coefficient or constant term without multiplying out every bracket.

Why combinations appear

How can the $64$ products hidden in $(1+x)^6$ collect into just seven terms, and why is the coefficient of $x^3$ exactly $20$ ? The answer is a counting problem wearing an algebra disguise.

You already know

Expanding brackets means choosing one term from each bracket and adding all the resulting products. ${}^nC_r$ counts the ways to choose $r$ positions from $n$ , when their order does not matter.

Start smaller: $(a+b)^3=(a+b)(a+b)(a+b)$ . Each bracket offers two choices, so there are $2^3=8$ products before collecting. The product $aab$ comes from choosing $b$ in the third bracket; $aba$ and $baa$ choose it in the second and first. All three are the same algebraic term, $a^2b$ . Choosing the positions matters to the count even though the letters then multiply to the same thing.

[DIAGRAM: asset_name: Choices behind a binomial expansion; asset_slug: binomial-choice-groups; caption: Eight choices from three brackets collect into four algebraic terms; description: Eight rows list aaa, aab, aba, baa, abb, bab, bba and bbb. The rows are grouped by the number of b choices, with group sizes 1, 3, 3 and 1, and arrows point to a cubed, 3 a squared b, 3 a b squared and b cubed. The b choices are sage and the collected terms are blue]
Eight choices from three brackets collect into four algebraic terms

This gives $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ . For $a^{n-r}b^r$ , choose the $r$ brackets supplying $b$ ; all other brackets supply $a$ . There are ${}^nC_r$ choices. This explains the coefficient rather than asking you to remember a triangle of numbers.

Binomial expansion

A binomial has two terms. Its binomial expansion, for a positive integer power, is the result of multiplying the repeated brackets and collecting like terms. For example, $(1+x)^3=1+3x+3x^2+x^3$ . The expression $(1+x+x^2)^3$ has three terms inside its bracket, so the binomial formula does not apply directly.

Here is a useful surprise: the coefficients of $(1+x)^n$ add to $2^n$ . Substituting $x=1$ shows why: the whole expression becomes $(1+1)^n$ . So $1+3+3+1=8$ is no coincidence. This coefficient-sum check is a consequence of the expansion, not a separate result on the formulae sheet.

Key idea

The coefficient ${}^nC_r$ counts which $r$ brackets supply the second term; each resulting product contains $n-r$ first terms and $r$ second terms.

Use that meaning to diagnose the coefficient in the next question before calculating anything.

Expanding positive integer powers

The counting argument gives a compact formula for the whole expansion; your job is to keep the two terms, including their signs, intact.

Binomial expansion

\begin{aligned} (a+b)^n={}&a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\cdots\\ &+{}^nC_ra^{n-r}b^r+\cdots+b^n \end{aligned}

Here $n$ is a positive integer and $r=0, 1, \ldots, n$ . Also ${}^nC_r=\dfrac{n!}{r!(n-r)!}$ . There are $n+1$ terms before any special values cause terms to vanish.
On the formulae sheet.

In every term the two exponents add to $n$ . To write an expansion in ascending powers of $x$ , start with the constant term, then $x$ , $x^2$ , and so on. Treat a term such as $2x$ as one complete choice, so its power is $(2x)^r$ .

Worked example 1: an expansion in ascending powers

Question. Expand $(1+2x)^4$ in ascending powers of $x$ .

Plan. Use $a=1$ , $b=2x$ and the coefficients $1, 4, 6, 4, 1$ , then simplify each power.

\begin{aligned} (1+2x)^4&=1+4(2x)+6(2x)^2+4(2x)^3+(2x)^4 && \text{binomial formula}\\ &=1+8x+24x^2+32x^3+16x^4 && \text{raise the whole term} \end{aligned}

Why these powers? Choosing $2x$ from $r$ brackets contributes $2^rx^r$ , so the number $2$ must be raised to the power too.

Answer. $1+8x+24x^2+32x^3+16x^4$ .

Check. At $x=1$ , the bracket gives $3^4=81$ , and the coefficients add to $1+8+24+32+16=81$ .

The next example changes just the sign of the second term. Odd powers now make negative terms; even powers stay positive.

Worked example 2: a negative second term

Question. Expand $(1-2x)^4$ in ascending powers of $x$ .

Plan. Use the same coefficients, but substitute $b=-2x$ before evaluating the powers.

\begin{aligned} (1-2x)^4&=1+4(-2x)+6(-2x)^2+4(-2x)^3+(-2x)^4 && \text{keep the minus sign}\\ &=1-8x+24x^2-32x^3+16x^4 && \text{odd powers are negative} \end{aligned}

Why put the minus inside the power? $(-2x)^2=4x^2$ , whereas $-(2x)^2=-4x^2$ . They are different expressions.

Answer. $1-8x+24x^2-32x^3+16x^4$ .

Check. At $x=1$ , $(-1)^4=1$ , and $1-8+24-32+16=1$ .

Both the numerical factor and the sign belong inside the powered bracket.

Watch out

Wrong: $(-3x)^3=-3x^3$ . The cube applies to the $3$ as well as to $x$ . Right: $(-3x)^3=(-3)^3x^3=-27x^3$ .

Now change the first term as well: its powers also need to appear, even though they contain no $x$ .

Finding a requested coefficient

When only one coefficient is requested, write only the term that can contain that power; a full expansion creates more opportunities for a slip.

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EN1 - The binomial expansion

Why combinations appear

Expanding positive integer powers

Finding a requested coefficient

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