P2 4.5 - Binomial expansion | Edexcel A-Level International Mathematics

Expand a positive-integer power without multiplying every bracket separately. You will use binomial coefficients to find particular terms, then track the contributions when an expansion is multiplied by another bracket.

Factorials and binomial coefficients

Why is the coefficient of $x^2$ in $(1+x)^4$ equal to $6$ ? There are four brackets, and an $x^2$ term comes from choosing $x$ in exactly two of them. The pairs of bracket positions are $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$ : six choices, each producing the same term.

The same counting pattern works for any positive-integer number of brackets. You need not list all the choices; factorials give a short calculation.

Factorial

For a positive integer $n$ , $n$ factorial, written $n!$ , is the product of the integers from $1$ to $n$ .

5!=5\times4\times3\times2\times1=120,\qquad 0!=1.

Choosing $r$ positions from $n$ produces a binomial coefficient. Its notation records the number of brackets and how many contribute the second term.

Binomial coefficient

For integers $0\le r\le n$ ,

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

In particular, $\binom n0=\binom nn=1$ .

In the formula booklet.

The denominator removes repeated arrangements of the same chosen positions. For instance, choosing positions $1$ then $2$ gives the same pair as choosing $2$ then $1$ . This is why choosing two positions gives $n(n-1)/2$ , rather than $n(n-1)$ .

Worked example 1: cancelling factorials

Question. Find $\binom72$ .

Plan. Cancel the common $5!$ before multiplying.

\binom72=\frac{7!}{2!5!} =\frac{7\times6\times5!}{2\times1\times5!} =21.

Answer. $\binom72=21$ .

Check. There are $7\times6=42$ ordered choices of two distinct positions; each unordered pair occurs twice, giving $42/2=21$ .

Change only the number selected from $2$ to $5$ . Selecting five positions leaves two unselected, so the coefficient should be unchanged.

Worked example 2: complementary selections

Question. Find $\binom75$ .

Plan. Use the same factorial formula; the two denominator factorials exchange places.

\binom75=\frac{7!}{5!2!} =\frac{7\times6}{2\times1} =21.

Answer. $\binom75=21$ .

Check. Every choice of five positions corresponds to exactly one omitted pair, confirming $\binom75=\binom72$ .

This symmetry, $\binom nr=\binom n{n-r}$ , is useful for checking calculator entries. Factorial notation is multiplication: $5!$ is not $5\times4$ alone.

Expanding a binomial

A binomial has two terms. In $(a+bx)^n$ , obtaining a term in $x^r$ means choosing $bx$ from $r$ of the $n$ brackets and $a$ from the other $n-r$ .

Each choice contributes $a^{n-r}(bx)^r$ , and there are $\binom nr$ choices. This explains both the coefficient and the opposing powers: the exponent on $a$ decreases as the exponent on $bx$ increases.

Positive-integer binomial expansion

For a positive integer $n$ ,

(a+bx)^n=\sum_{r=0}^{n}\binom nr a^{n-r}(bx)^r.

Its beginning is

a^n+na^{n-1}bx+\binom n2a^{n-2}b^2x^2+\binom n3a^{n-3}b^3x^3+\cdots.

There are $n+1$ terms, ending with $b^nx^n$ . The complete polynomial is exact for every real $x$ .

The expansion and coefficient formula are in the formula booklet.

“Ascending powers of $x$ ” means constant term first, then $x,x^2,x^3,\ldots$ . If only the first four terms are requested, stop after $x^3$ ; an ellipsis indicates that higher powers remain.

Worked example 3: a non-unit constant

Question. Find the first four terms, in ascending powers of $x$ , of $(2+x)^5$ , in simplest form.

Plan. Use powers $0,1,2,3$ of the entire second term $x$ , with powers $5,4,3,2$ of $2$ .

\begin{aligned} (2+x)^5 &=2^5+5(2^4)x+10(2^3)x^2+10(2^2)x^3+\cdots\\ &=32+80x+80x^2+40x^3+\cdots \end{aligned}

Answer. $32+80x+80x^2+40x^3$ .

Check. At $x=1$ , the complete expansion must total $3^5=243$ . The omitted terms are $10x^4+x^5$ , and $32+80+80+40+10+1=243$ .

Change only the sign of $x$ . Odd powers change sign; even powers do not.

Worked example 4: a negative second term

Question. Find the first four terms, in ascending powers of $x$ , of $(2-x)^5$ , in simplest form.

Plan. Use $-x$ as the entire second term, including its sign.

\begin{aligned} (2-x)^5 &=2^5+5(2^4)(-x)+10(2^3)(-x)^2+10(2^2)(-x)^3+\cdots\\ &=32-80x+80x^2-40x^3+\cdots \end{aligned}

Answer. $32-80x+80x^2-40x^3$ .

Check. The omitted terms are $10x^4-x^5$ . At $x=1$ , $32-80+80-40+10-1=1=(2-1)^5$ .

The same expansion can approximate a nearby numerical power. For example, to use $(2-x)^5$ for $1.98^5$ , first solve $2-x=1.98$ , giving $x=0.02$ . The first four terms give an approximation because two terms have been omitted; the complete expansion gives the exact value.

Watch out

The square of a negative second term needs brackets: $(-3x)^2=9x^2$ , whereas $-3x^2$ is a different expression. Apply the exponent to the coefficient, sign and variable together.

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P2 4.5 - Binomial expansion

Factorials and binomial coefficients

Expanding a binomial

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