Binomial expansion for rational nAQA A-Level Maths: Revision notes

Section 1

The expansion of (1 + x)ⁿ for any rational n

For a positive integer $n$ the expansion of $(1+x)^n$ stops. For any rational $n$ (negative or fractional) it goes on forever: $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\dots$ This is an infinite series and it is only valid when $|x|<1$ . Each coefficient takes one more factor on the top and one more factorial below. Keep brackets round negative numbers: for $n=-2$ the $x^2$ coefficient is $\frac{(-2)(-3)}{2!}=3$ .

Key termsrational ninfinite series

Common mistake

Dropping a minus sign when $n$ is negative. Write every factor in brackets, e.g. $(-2)(-3)$ .

Section 2

Expanding (a + bx)ⁿ

The formula needs a bracket that starts with $1$ , so take out the first term: $(a+bx)^n=a^n\left(1+\frac{b}{a}x\right)^n.$ Expand with $\frac{b}{a}x$ in place of $x$ , then multiply every term by $a^n$ . The expansion is valid when $\left|\frac{bx}{a}\right|<1$ , i.e. $|x|<\left|\frac{a}{b}\right|$ (proof is not required). Example: $\sqrt{4+x}=2\left(1+\frac{x}{4}\right)^{\frac12}=2\left[1+\frac{x}{8}-\frac{x^2}{128}+\dots\right]=2+\frac{x}{4}-\frac{x^2}{64}+\dots$ , valid for $|x|<4$ .

Key termsvalidity

Common mistake

Forgetting to raise $a$ to the power $n$ . Taking $4$ out of $\sqrt{4+x}$ gives $4^{\frac12}=2$ , not $4$ .

Exam tip

Write the factor out first, then a bracket beginning with $1$ , then the validity condition, before expanding.

Section 3

Worked example: a negative power

Expand $(1+3x)^{-2}$ up to $x^2$ . Here $n=-2$ and the variable is $3x$ : $1+(-2)(3x)+\frac{(-2)(-3)}{2!}(3x)^2=1-6x+27x^2.$ Validity: $|3x|<1$ , so $|x|<\frac13$ . The next term is $\frac{(-2)(-3)(-4)}{3!}(3x)^3=-108x^3$ . Note that $(3x)^2=9x^2$ : square the $3$ as well as the $x$ .

Exam tip

For $(1+bx)^{-2}$ with $b>0$ the signs alternate. A pattern that does not alternate is a warning.

Section 4

Using the expansion for approximation

Substitute a value of $x$ that is inside the validity range and small, so the terms shrink quickly. To estimate $\frac{1}{2.5}$ use $\frac{1}{2+5x}=\frac12-\frac{5x}{4}+\frac{25x^2}{8}+\dots$ with $x=0.1$ (valid, since $|x|<0.4$ ): $0.5-0.125+0.03125=0.40625$ . The exact value is $0.4$ , so the percentage error is $\frac{0.40625-0.4}{0.4}\times100=1.56\%$ . More terms give a better estimate. Choose $x$ so that the bracket equals the number you want: for $1.03^{-2}$ write $1.03=1+3(0.01)$ and use $x=0.01$ .

Key termspercentage error

Common mistake

Using an $x$ outside the validity range. The series then does not converge to the function, however many terms you take.

Section 5

Products and finding a surd

For $\frac{1+x}{\sqrt{1-2x}}$ write $(1+x)(1-2x)^{-\frac12}$ , expand the second factor and multiply out, collecting like powers: $(1-2x)^{-\frac12}=1+x+\frac32x^2+\frac52x^3+\dots,\qquad(1+x)(\dots)=1+2x+\frac52x^2+4x^3+\dots,\quad|x|<\tfrac12.$ Put $x=0.01$ into both sides: $\frac{1.01}{\sqrt{0.98}}=\frac{101}{70\sqrt2}$ exactly, and the series gives $1.020254$ . So $\sqrt2\approx\frac{101}{70\times1.020254}=1.4142$ . The validity range of a product is that of the most restrictive factor.

Exam tip

Use exact surd forms such as $\sqrt{0.98}=0.7\sqrt2$ to link the series to the number you are estimating.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Binomial expansion for rational n

The function $f(x)=(1+3x)^{-2}$ is expanded in ascending powers of $x$ .
Use the first three terms of the expansion, with a suitable value of $x$ , to estimate $1.03^{-2}$ .2 marks
Let $g(x)=\sqrt{4+x}$ .
Find the first three terms of the expansion of $g(x)$ in ascending powers of $x$ .2 marks
Let $f(x)=\frac{1}{2+5x}$ .
Find the expansion of $f(x)$ in ascending powers of $x$ , up to and including the term in $x^2$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).