Pure 4.1 - Binomial expansion | Edexcel A-Level Mathematics

A bracket raised to a power can hide a long polynomial, a square root or a reciprocal. You will use binomial coefficients to uncover its terms, extend the method to rational powers, and decide when a truncated expansion gives a sensible approximation. You will also combine expansions after partial-fraction decomposition.

Positive integer powers and coefficients

How can a few fractions give a useful approximation to $\sqrt3$ ? We will get there by first understanding something simpler: why does expanding $(1+x)^3$ produce $1+3x+3x^2+x^3$ ?

You already know

Multiplying brackets means choosing one term from each bracket and adding the resulting products. Index laws give $(bx)^r=b^rx^r$ , including the sign of $b$ .

In three brackets, there are three ways to choose exactly one $x$ , and three ways to choose exactly two. Those counts create the two coefficients $3$ .

Binomial coefficient

For integers $n\ge0$ and $0\le r\le n$ ,

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

counts the ways to choose $r$ objects from $n$ . Here $n!=n(n-1)\cdots2\cdot1$ and $0!=1$ . For example, $\binom42=6$ ; $\binom45$ is outside this counting definition.

The same count appears in binomial probability: it counts which $r$ of $n$ trials are successes. In algebra, it counts which brackets supply the second term. Pascal's triangle builds these counts without factorials.

$n$	Coefficients of $(p+q)^n$
$0$	$1$
$1$	$1,1$
$2$	$1,2,1$
$3$	$1,3,3,1$
$4$	$1,4,6,4,1$
$5$	$1,5,10,10,5,1$

Each inner entry is the sum of the two above it: $\binom nr=\binom{n-1}{r-1}+\binom{n-1}r$ . Choosing $r$ objects is equivalent to choosing the $n-r$ objects left out, so $\binom nr=\binom n{n-r}$ . This explains the symmetry.

Positive integer binomial expansion

For a positive integer $n$ ,

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

The sum means add the terms for $r=0,1,\ldots,n$ . It stops after $n+1$ terms and is valid for every real $x$ .

In the formula booklet.

An unexpected consequence is that the entries in row $n$ add to $2^n$ : put $p=q=1$ . The row $1,4,6,4,1$ therefore totals $16$ , not an accidental pattern.

Worked example 1: keeping the negative term in brackets

Question. Expand $(2-3x)^4$ .

Plan. Use the fourth Pascal row, keeping $-3x$ together in every power.

\begin{aligned} (2-3x)^4&=2^4+4(2^3)(-3x)+6(2^2)(-3x)^2\\ &\quad+4(2)(-3x)^3+(-3x)^4 &&\text{choose each power}\\ &=16-96x+216x^2-216x^3+81x^4 &&\text{simplify} \end{aligned}

Why this step? Bracketing the whole negative term makes the signs alternate correctly.

Answer. $16-96x+216x^2-216x^3+81x^4$ .

Check. At $x=1$ , the bracket gives $(-1)^4=1$ , and the polynomial gives $16-96+216-216+81=1$ .

Change only the sign of the second term. The coefficients have the same magnitudes, but all the terms are now positive.

Worked example 2: a positive second term

Question. Expand $(2+3x)^4$ .

Plan. Use the same row with $3x$ in place of $-3x$ .

\begin{aligned} (2+3x)^4&=2^4+4(2^3)(3x)+6(2^2)(3x)^2\\ &\quad+4(2)(3x)^3+(3x)^4 &&\text{binomial theorem}\\ &=16+96x+216x^2+216x^3+81x^4 &&\text{collect powers} \end{aligned}

Why this step? The coefficient of $x^2$ includes $3^2$ , as well as the Pascal entry $6$ .

Answer. $16+96x+216x^2+216x^3+81x^4$ .

Check. At $x=1$ , the coefficients total $625=5^4$ .

The Pascal row counts choices; it does not replace the powers of the bracket's terms.

Watch out

The line $(2+3x)^4=16+4x+6x^2+4x^3+x^4$ is wrong. Each coefficient also needs the appropriate powers of $2$ and $3$ .

Try the same expansion with a new second coefficient.

Finding a particular term efficiently

If you want one coefficient, expanding everything creates unnecessary work. For example, in $(5-2x)^6$ , only the choice of three second terms can produce $x^3$ .

A specified term

For positive integer $n$ , the term containing $x^r$ in $(a+bx)^n$ is

T_{r+1}=\binom nr a^{n-r}b^rx^r\qquad(0\le r\le n)

Its coefficient is $\binom nr a^{n-r}b^r$ . This is the corresponding term of the supplied binomial theorem.

In the formula booklet.

Term numbers start at $1$ , while powers start at $0$ . The fourth term contains $x^3$ , not $x^4$ .

Worked example 3: finding a term without expanding

Question. Find the term in $x^3$ in $(5-2x)^6$ .

Plan. Set $r=3$ in the general term.

\begin{aligned} T_4&=\binom63 5^{6-3}(-2x)^3 &&\text{select power three}\\ &=20(125)(-8)x^3 &&\text{evaluate factors}\\ &=-20000x^3 &&\text{multiply} \end{aligned}

Why this step? Choosing $r=3$ selects three factors containing $x$ .

Answer. The term is $-20000x^3$ ; its coefficient is $-20000$ .

Check. An odd power of $-2$ makes the coefficient negative.

Now change only the requested power. This also changes which power of $5$ remains.

Worked example 4: changing the requested coefficient

Question. Find the coefficient of $x^4$ in $(5-2x)^6$ .

Plan. Use $r=4$ , then omit $x^4$ from the final coefficient.

\begin{aligned} T_5&=\binom64 5^2(-2x)^4 &&\text{general term}\\ &=15(25)(16)x^4 &&\text{even power}\\ &=6000x^4 &&\text{multiply} \end{aligned}

Why this step? A coefficient is the numerical multiplier, so the final answer does not include $x^4$ .

Answer. $6000$ .

Check. The powers chosen from the brackets total $2+4=6$ .

The same method lets you form an equation if a coefficient contains an unknown parameter.

Watch out

Using $r=5$ for the term in $x^4$ selects the wrong power. The fifth term uses $r=4$ , because the constant is the first term.

Keep the requested power fixed and change the bracket's numbers.

Rational powers and infinite series

A square root cannot usually equal a finite polynomial. Instead, $(1+u)^{1/2}$ begins $1+\frac12u-\frac18u^2+\cdots$ , with further terms continuing indefinitely.

Rational binomial expansion

For rational $n$ outside the non-negative integers,

(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\frac{n(n-1)(n-2)}{3!}u^3+\cdots

Use this expansion when $|u|<1$ . This sufficient range is the one required here; endpoint behaviour needs separate analysis and is not assumed.

In the formula booklet.

To expand $(a+bx)^n$ about $x=0$ , first write $a^n(1+\frac ba x)^n$ . Take $a>0$ for a real non-integer power, so this factorisation stays on the usual real branch. A negative integer power needs $a\ne0$ . The whole inner expression replaces $u$ .

Worked example 5: factoring before a square-root expansion

Question. Expand $(4-3x)^{1/2}$ through $x^3$ .

Plan. Factor out $4$ , then use $n=\frac12$ and $u=-\frac34x$ .

\begin{aligned} (4-3x)^{1/2}&=2\left(1-\frac34x\right)^{1/2} &&\text{factor constant}\\ &=2\left[1+\frac12\left(-\frac34x\right)-\frac18\left(-\frac34x\right)^2+\frac1{16}\left(-\frac34x\right)^3+\cdots\right] &&\text{substitute whole }u\\ &=2-\frac34x-\frac9{64}x^2-\frac{27}{512}x^3+\cdots &&\text{multiply by two} \end{aligned}

Why this step? The supplied series starts with $1$ inside the bracket, so the outside factor $2$ is essential.

Answer. $2-\frac34x-\frac9{64}x^2-\frac{27}{512}x^3+\cdots$ , for $|x|<\frac43$ .

Check. Squaring the displayed polynomial gives $4-3x$ with zero coefficients of $x^2$ and $x^3$ , as required.

Change only the exponent, from $\frac12$ to $-\frac12$ . The outer factor and the signs must both be recomputed.

Worked example 6: a reciprocal square root

Question. Expand $(4-3x)^{-1/2}$ through $x^3$ .

Plan. Use $4^{-1/2}=\frac12$ and the coefficients $1,-\frac12,\frac38,-\frac5{16}$ .

\begin{aligned} (4-3x)^{-1/2}&=\frac12\left(1-\frac34x\right)^{-1/2} &&\text{factor constant}\\ &=\frac12\left[1-\frac12\left(-\frac34x\right)+\frac38\left(-\frac34x\right)^2-\frac5{16}\left(-\frac34x\right)^3+\cdots\right] &&\text{binomial coefficients}\\ &=\frac12+\frac3{16}x+\frac{27}{256}x^2+\frac{135}{2048}x^3+\cdots &&\text{simplify} \end{aligned}

Why this step? The sign belongs to $u$ as well as to the coefficient $-\frac12$ , so the linear term is positive.

Answer. $\frac12+\frac3{16}x+\frac{27}{256}x^2+\frac{135}{2048}x^3+\cdots$ , for $|x|<\frac43$ .

Check. Multiplying this series by the square-root series in Example 5 gives constant $1$ and zero coefficients of $x$ , $x^2$ and $x^3$ , as reciprocal factors should.

The coefficient of $u^2$ is $\frac{n(n-1)}2$ ; it is not $\frac{n^2}2$ .

Watch out

Factoring $(4-3x)^{-1/2}$ gives an outside factor $\frac12$ , not $2$ . Apply the original exponent to the constant too.

Use new constants with the same reciprocal-root exponent.

Validity and approximation

For $\sqrt{4-x}$ , choosing $x=1$ produces the $\sqrt3$ in our opening question. Here the small quantity is $-\frac14$ , not $1$ ; that distinction makes the approximation work.

Validity after factoring

For the infinite binomial expansion about $x=0$ , write

(a+bx)^n=a^n\left(1+\frac ba x\right)^n

with the real-power conditions stated above. Its required validity condition is

\left|\frac{bx}{a}\right|<1

For $b\ne0$ , this gives $|x|<|a/b|$ . The general $|u|<1$ condition is supplied; you must identify $u$ yourself.

Not in the formula booklet: learn it.

Validity and accuracy are different. Inside the interval, the full series represents the function; a short truncation is still only approximate. Smaller $|u|$ usually makes the first few terms more useful. A positive integer power gives an exact finite polynomial, with no convergence restriction. For exponent $0$ , the expression is the constant $1$ wherever the original power is defined; it does not require an infinite expansion.

Worked example 7: approximating a square root

Question. Use the first four terms of $\sqrt{4-x}$ to approximate $\sqrt3$ .

Plan. Expand in powers of $-x/4$ , check $x=1$ , then substitute.

\begin{aligned} \sqrt{4-x}&=2\left(1-\frac x4\right)^{1/2} &&\text{factor four}\\ &=2-\frac x4-\frac{x^2}{64}-\frac{x^3}{512}+\cdots &&\text{binomial expansion}\\ \sqrt3&\approx2-\frac14-\frac1{64}-\frac1{512} &&\text{put }x=1\\ &=\frac{887}{512}=1.732421875 &&\text{evaluate} \end{aligned}

Why this step? $|-1/4|<1$ , so the substitution lies within $-4<x<4$ .

Answer. $\sqrt3\approx1.732421875$ using the stated truncation.

Check. The calculator value is $1.7320508\ldots$ ; the absolute error is about $0.000371$ .

Change only the target, from $\sqrt3$ to $\sqrt{3.8}$ . The inner quantity is now smaller, so expect a better approximation from the same number of terms.

Worked example 8: a smaller substitution

Question. Use the same four terms to approximate $\sqrt{3.8}$ .

Plan. Set $x=0.2$ and check the inner expression.

\begin{aligned} \left|-\frac{0.2}{4}\right|&=0.05<1 &&\text{valid substitution}\\ \sqrt{3.8}&\approx2-\frac{0.2}{4}-\frac{0.2^2}{64}-\frac{0.2^3}{512} &&\text{same truncation}\\ &=1.949359375 &&\text{evaluate} \end{aligned}

Why this step? We solve $4-x=3.8$ ; using $x=3.8$ would approximate a different root.

Answer. $1.949359375$ .

Check. The calculator gives $1.949358869\ldots$ , an error of about $0.000000506$ .

A calculator check measures this particular error; it does not prove a general error bound.

Watch out

The equation $\sqrt{4-x}=2-\frac x4-\frac{x^2}{64}-\frac{x^3}{512}$ is false in general. Use $\approx$ after omitting the remaining terms, or retain $+\cdots$ for the full series.

Choose another target, and justify the substitution before calculating.

Expanding rational functions after decomposition

A complicated denominator may hide two easy reciprocal expansions. Partial fractions expose them. Both expansions must be valid at the chosen $x$ , so their validity intervals must overlap.

You already know

For distinct linear factors, $\frac{px+q}{(2+x)(1-2x)}=\frac A{2+x}+\frac B{1-2x}$ . Multiplying through by the denominator gives an identity you can use to find $A$ and $B$ .

The exponent $-1$ is a rational exponent. Its binomial series is $(1+u)^{-1}=1-u+u^2-\cdots$ , valid for $|u|<1$ .

Worked example 9: decomposing before expanding

Question. Expand $\frac{-1-13x}{(2+x)(1-2x)}$ through $x^2$ , and state its validity interval.

Plan. Decompose, expand each reciprocal, then intersect the ranges.

\begin{aligned} -1-13x&=A(1-2x)+B(2+x) &&\text{clear denominator}\\ A+2B&=-1,\quad -2A+B=-13 &&\text{compare coefficients}\\ A&=5,\quad B=-3 &&\text{solve}\\ \frac5{2+x}&=\frac52\left(1-\frac x2+\frac{x^2}4+\cdots\right) &&|x|<2\\ \frac{-3}{1-2x}&=-3(1+2x+4x^2+\cdots) &&|x|<\frac12 \end{aligned}

Why this step? A product denominator does not permit adding separate reciprocal expansions until the partial-fraction identity has been established.

Answer. Adding like powers gives

-\frac12-\frac{29}4x-\frac{91}8x^2+\cdots,\qquad -\frac12<x<\frac12

Check. At $x=0$ , the function gives $-1/2$ , matching the constant term. Recombining the partial fractions gives the original numerator.

Change only the numerator. The two denominators, and hence the common validity interval, stay the same.

Worked example 10: the same poles with new coefficients

Question. Expand $\frac{8-x}{(2+x)(1-2x)}$ through $x^2$ , stating the interval.

Plan. Find new partial-fraction constants, using the same reciprocal series.

\begin{aligned} 8-x&=A(1-2x)+B(2+x) &&\text{identity}\\ A+2B&=8,\quad -2A+B=-1 &&\text{equate coefficients}\\ A&=2,\quad B=3 &&\text{solve}\\ \frac2{2+x}&=1-\frac x2+\frac{x^2}4+\cdots &&\text{first expansion}\\ \frac3{1-2x}&=3+6x+12x^2+\cdots &&\text{second expansion} \end{aligned}

Why this step? A new numerator changes the coefficients, but the two unchanged inner expressions still impose $|x|<2$ and $|x|<1/2$ .

Answer. $4+\frac{11}2x+\frac{49}4x^2+\cdots$ , for $-\frac12<x<\frac12$ .

Check. Multiplying the truncated series by $(2+x)(1-2x)=2-3x-2x^2$ gives $8-x$ with zero $x^2$ coefficient.

The narrower interval controls the combined expansion.

Watch out

Combining $|x|<2$ and $|x|<1/2$ gives $|x|<1/2$ , not $|x|<2$ . Both component series must converge at once.

Keep the denominators and change the numerator again.

Choosing the expansion and its validity range

Choosing the right expansion includes choosing which restrictions belong to which factor. In a product, a finite polynomial contributes no convergence restriction; an infinite series does. Terms of the same final degree can come from different factors, so collect every contribution.

Expression	Useful first step	Required range
$(a+bx)^m$ , non-negative integer $m$	finite theorem or one general term	all real $x$ for positive $m$ ; original domain for $m=0$
$(a+bx)^n$ , rational $n$ outside the non-negative integers	factor to $a^n(1+u)^n$	$
sum of partial fractions	expand each simple fraction	common interval
polynomial times an infinite series	retain all products of the requested degree	the infinite factor's interval

Worked example 11: combining a polynomial and an infinite expansion

Question. Find the coefficient of $x^2$ in $(1+2x)^2(1-x)^{-1/2}$ , and state the validity interval.

Plan. Expand each factor only through $x^2$ and collect products of total degree two.

\begin{aligned} (1+2x)^2&=1+4x+4x^2 &&\text{exact polynomial}\\ (1-x)^{-1/2}&=1+\frac12x+\frac38x^2+\cdots &&\text{binomial series}\\ [x^2]\text{ coefficient}&=\frac38+4\left(\frac12\right)+4 &&\text{degrees }0+2,1+1,2+0\\ &=\frac{51}8 &&\text{add contributions} \end{aligned}

Why this step? Multiplying just the two $x^2$ terms would produce $x^4$ , not $x^2$ .

Answer. The coefficient is $51/8$ , for $-1<x<1$ .

Check. The constant is $1$ , and the linear coefficient is $4+1/2=9/2$ , consistent with multiplying the first two terms.

Change only the coefficient inside the finite bracket. The rational-power expansion and its range stay fixed.

Worked example 12: collecting cross terms

Question. Find the coefficient of $x^2$ in $(1+3x)^2(1-x)^{-1/2}$ , and state the interval.

Plan. Replace the polynomial by $1+6x+9x^2$ .

\begin{aligned} (1+3x)^2&=1+6x+9x^2 &&\text{finite expansion}\\ [x^2]\text{ coefficient}&=\frac38+6\left(\frac12\right)+9 &&\text{collect total degree two}\\ &=\frac{99}8 &&\text{simplify} \end{aligned}

Why this step? The cross product $6x\times\frac12x$ is a separate contribution to the requested coefficient.

Answer. $99/8$ , with $-1<x<1$ .

Check. Increasing the bracket coefficient from $2$ to $3$ increases the coefficient by $(99-51)/8=6$ , matching the extra cross and quadratic contributions.

We can now answer the opening question: factoring $4-x$ turns the square root into a series in a small quantity. The first four terms give $887/512$ for $\sqrt3$ , and the validity check explains why that substitution is legitimate.

Exam tip

Show the inner expression and its inequality. Writing only “valid for small $x$ ” does not give the requested interval, and a truncated polynomial is an approximation rather than an exact identity.

Try one last change to the finite bracket.

Quick Recap

Binomial coefficients count choices; Pascal's triangle gives their symmetry and adjacent-sum pattern.
A specified coefficient uses the general term, with power $r$ in term $r+1$ .
For a rational power, factor to $a^n(1+u)^n$ , keep the outside factor, and require $|u|<1$ .
Combine validity intervals and collect every product of the requested degree.
Truncating an infinite expansion gives an approximation.

The opening square root is accessible because $\sqrt3=2(1-1/4)^{1/2}$ ; four terms give $1.732421875$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 4.1 - Binomial expansion

Positive integer powers and coefficients

Finding a particular term efficiently

Rational powers and infinite series

Validity and approximation

Expanding rational functions after decomposition

Choosing the expansion and its validity range

Quick Recap

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