6A - Binomial expansions with integer powers

Repeated brackets soon become slow to multiply out. You will see where binomial coefficients come from, use them to expand positive integer powers, and find one required term without calculating the rest.

Binomial coefficients

Why is the coefficient of $x^2$ in $(1+x)^5$ equal to $10$ , rather than $5$ ?

Think of five brackets $(1+x)(1+x)(1+x)(1+x)(1+x)$ . To make an $x^2$ term, choose $x$ from two brackets and $1$ from the other three. There are $5\times4$ ordered choices of two brackets, but each pair has been counted twice: choosing brackets 2 then 4 gives the same product as choosing 4 then 2. There are therefore $5\times4\div2=10$ copies of $x^2$ .

Factorial

For a positive integer $n$ , factorial means $n!=n(n-1)\cdots2\times1$ . For example, $4!=24$ . We define $0!=1$ , which makes the coefficient formula work at both ends of an expansion.

Choosing $r$ brackets in order counts the same selection $r!$ times. Dividing by $r!$ removes those repetitions. The result is a binomial coefficient, written $\binom nr$ , read “ $n$ choose $r$ ”.

Binomial coefficients

For positive integer $n$ and integer $r$ with $0\le r\le n$ ,

\binom nr=\frac{n!}{r!(n-r)!}.

For $r\ge1$ , cancellation also gives

\binom nr=\frac{n(n-1)\cdots(n-r+1)}{r!}.

The general binomial series is on the formulae sheet; this compact coefficient notation is not printed separately.

The two end coefficients are $\binom n0=\binom nn=1$ : choose none of the brackets, or all of them. Choosing $r$ brackets is equivalent to identifying the $n-r$ brackets left behind, so $\binom nr=\binom n{n-r}$ .

Worked example 1: counting a quadratic coefficient

Question. Find $\binom52$ , the coefficient of $x^2$ in $(1+x)^5$ .

Plan. Use the factorial formula and cancel before multiplying.

\begin{aligned} \binom52&=\frac{5!}{2!3!} &&\text{substitute }n=5\text{ and }r=2\\ &=\frac{5\times4\times3!}{2\times1\times3!} &&\text{expose the common factor}\\ &=10 &&\text{cancel }3!\text{ and simplify} \end{aligned}

Answer. The coefficient is $10$ .

Check. Listing pairs of bracket positions gives $4+3+2+1=10$ pairs, independently confirming the count.

Now increase the number of brackets by one, keeping the required power $x^2$ . Only the top value in the coefficient changes.

Worked example 2: one more bracket

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

Plan. Choose the two brackets that supply $x$ .

\begin{aligned} \binom62&=\frac{6!}{2!4!} &&\text{choose two from six}\\ &=\frac{6\times5}{2\times1} &&\text{cancel }4!\\ &=15 &&\text{evaluate} \end{aligned}

Answer. The coefficient is $15$ .

Check. Of the pairs, $10$ avoid the new sixth bracket and $5$ include it. Their total is $15$ .

Factorials multiply consecutive positive integers; they are not powers of one number.

Watch out

$4!=4\times3\times2\times1=24$ , whereas $4^4=256$ . Also, $0!=1$ , not $0$ ; otherwise the end coefficients would involve division by zero.

Use the same counting idea for the next coefficient.

Expanding a binomial

A binomial has two terms, and its positive integer power expands into a finite sum.

For $(a+bx)^n$ , choosing $bx$ from $r$ brackets leaves $a$ from $n-r$ brackets. Each such product is $a^{n-r}(bx)^r$ , and there are $\binom nr$ of them. This explains both the coefficient and the powers.

Positive integer binomial expansion

For positive integer $n$ ,

(a+bx)^n=a^n+na^{n-1}(bx)+\frac{n(n-1)}{2!}a^{n-2}(bx)^2+\cdots+(bx)^n.

The term containing $(bx)^r$ is $\binom nr a^{n-r}(bx)^r$ . There are $n+1$ terms before any zero coefficients are removed. This is a polynomial identity, valid for every real $x$ .

Write coefficients first, then let the power of the first term decrease while the power of the second increases. For a cube the coefficients are $1,3,3,1$ . Ascending powers means writing the constant term first, then the $x$ , $x^2$ and higher-power terms.

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6A - Binomial expansions with integer powers

Binomial coefficients

Expanding a binomial

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