Mathematics 9709 · Pure Mathematics 3

Binomial expansion of (a + x)ⁿ

When the bracket starts with $a$ instead of 1, pull the a out first: $(a+x)^n=a^n\big(1+\tfrac{x}{a}\big)^n$ .
Now it is the same job as §7.3: expand $(1+\tfrac{x}{a})^n$ , then multiply every term by $a^n$ .

$(a+x)^n=a^n\Big[1+n\tfrac{x}{a}+\tfrac{n(n-1)}{2!}\big(\tfrac{x}{a}\big)^2+\cdots\Big]$ , valid when $|x|<|a|$ . Two things people forget: the $a^n$ out front, and reading the validity off $\big|\tfrac{x}{a}\big|<1$ .

Factor out aⁿ, then expand

$(a+x)^n=a^n\Big[\,1+n\tfrac{x}{a}+\tfrac{n(n-1)}{2!}\big(\tfrac{x}{a}\big)^2+\cdots\Big],\quad \left|\tfrac{x}{a}\right|<1$

The §7.3 formula needs a $1$ at the front, so pull $a^n$ outside to make one.
Once you have expanded the bracket to the term you need, multiply every term by $a^n$ .

Why a negative or fractional a needs care

$a^n$ can itself be a fraction or a surd. For $(2+x)^{-3}$ the front factor is $2^{-3}=\tfrac18$ ; for $(4+x)^{-1/2}$ it is $4^{-1/2}=\tfrac12$ . Work that number out carefully before you multiply through. The validity $|x|<|a|$ uses the size of $a$ , so it is always a positive number.

Worked examples

Worked example

Textbook: expand $(2+x)^{-3}$ up to $x^3$ , with validity

Factor out the 2: $(2+x)^{-3}=2^{-3}\big(1+\tfrac{x}{2}\big)^{-3}=\tfrac18\big(1+\tfrac{x}{2}\big)^{-3}$ .
Expand the bracket with $n=-3$ , inside $\tfrac{x}{2}$ : $1-3\tfrac{x}{2}+6\big(\tfrac{x}{2}\big)^2-10\big(\tfrac{x}{2}\big)^3=1-\tfrac{3x}{2}+\tfrac{3x^2}{2}-\tfrac{5x^3}{4}$ .
Multiply through by $\tfrac18$ , and read validity from $\big|\tfrac{x}{2}\big|<1$ , i.e. $|x|<2$ .

Answer

\tfrac18-\tfrac{3x}{16}+\tfrac{3x^2}{16}-\tfrac{5x^3}{32}+\cdots,\quad |x|<2

Worked example

A surd denominator: expand $\dfrac{1}{\sqrt{4+x}}$ up to $x^2$ , with validity

Rewrite as a power: $\dfrac{1}{\sqrt{4+x}}=(4+x)^{-1/2}$ , then factor: $=4^{-1/2}\big(1+\tfrac{x}{4}\big)^{-1/2}=\tfrac12\big(1+\tfrac{x}{4}\big)^{-1/2}$ .
Expand with $n=-\tfrac12$ , inside $\tfrac{x}{4}$ : $1-\tfrac12\tfrac{x}{4}+\tfrac{(-\frac12)(-\frac32)}{2}\big(\tfrac{x}{4}\big)^2=1-\tfrac{x}{8}+\tfrac{3x^2}{128}$ .
Multiply by $\tfrac12$ ; validity is $\big|\tfrac{x}{4}\big|<1$ , i.e. $|x|<4$ .

Answer

\tfrac12-\tfrac{x}{16}+\tfrac{3x^2}{256}+\cdots,\quad |x|<4

Worked example

First 3 terms of $(1+ax)^n$ are $1-24x+384x^2$ . Find $a$ and $n$

Match terms: the $x$ term gives $na=-24$ ; the $x^2$ term gives $\tfrac{n(n-1)}{2}a^2=384$ .
From the first, $a^2=\tfrac{576}{n^2}$ . Substitute: $\tfrac{n(n-1)}{2}\cdot\tfrac{576}{n^2}=384\Rightarrow 576(n-1)=768n\Rightarrow n=-3$ .
Then $a=\tfrac{-24}{n}=8$ .

Answer

n=-3,\quad a=8

Where the marks go

Common mistake

Don't forget to multiply the whole bracket expansion by

a^n

, and read the validity off the inside:

(a+x)^n

needs

|x|<|a|

, not

|x|<1

Common mistake

A reciprocal surd like

\tfrac1{\sqrt{4+x}}

is really a power:

(4+x)^{-1/2}

. Turn it into a single power before you do anything else. The binomial only works once it is written as one power.

Now you try

Expand $(2-3x)^{-2}$ in ascending powers of $x$ up to $x^2$ , simplify the coefficients. (9709/31 May/Jun 2020 Q2)

Your turn— tap to reveal the worked answer (9709/31 May/Jun 2020 Q2(a))

Pull the 2 out: $(2-3x)^{-2}=2^{-2}\big(1-\tfrac{3x}{2}\big)^{-2}=\tfrac14\big(1-\tfrac{3x}{2}\big)^{-2}$ . With $n=-2$ and inside $-\tfrac{3x}{2}$ :

$\tfrac14\Big[1+(-2)\big(-\tfrac{3x}{2}\big)+\tfrac{(-2)(-3)}{2}\big(-\tfrac{3x}{2}\big)^2\Big]=\tfrac14\big[1+3x+\tfrac{27}{4}x^2\big]$

Answer

\tfrac14+\tfrac{3x}{4}+\tfrac{27x^2}{16}+\cdots

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Binomial expansion of (a + x)ⁿ

When the bracket starts with $a$ instead of 1, pull the a out first: $(a+x)^n=a^n\big(1+\tfrac{x}{a}\big)^n$ .
Now it is the same job as §7.3: expand $(1+\tfrac{x}{a})^n$ , then multiply every term by $a^n$ .

Factor out aⁿ, then expand

$(a+x)^n=a^n\Big[\,1+n\tfrac{x}{a}+\tfrac{n(n-1)}{2!}\big(\tfrac{x}{a}\big)^2+\cdots\Big],\quad \left|\tfrac{x}{a}\right|<1$

The §7.3 formula needs a $1$ at the front, so pull $a^n$ outside to make one.
Once you have expanded the bracket to the term you need, multiply every term by $a^n$ .

Why a negative or fractional a needs care

Worked examples

Worked example

Textbook: expand $(2+x)^{-3}$ up to $x^3$ , with validity

Factor out the 2: $(2+x)^{-3}=2^{-3}\big(1+\tfrac{x}{2}\big)^{-3}=\tfrac18\big(1+\tfrac{x}{2}\big)^{-3}$ .
Expand the bracket with $n=-3$ , inside $\tfrac{x}{2}$ : $1-3\tfrac{x}{2}+6\big(\tfrac{x}{2}\big)^2-10\big(\tfrac{x}{2}\big)^3=1-\tfrac{3x}{2}+\tfrac{3x^2}{2}-\tfrac{5x^3}{4}$ .
Multiply through by $\tfrac18$ , and read validity from $\big|\tfrac{x}{2}\big|<1$ , i.e. $|x|<2$ .

Answer

\tfrac18-\tfrac{3x}{16}+\tfrac{3x^2}{16}-\tfrac{5x^3}{32}+\cdots,\quad |x|<2

Worked example

A surd denominator: expand $\dfrac{1}{\sqrt{4+x}}$ up to $x^2$ , with validity

Rewrite as a power: $\dfrac{1}{\sqrt{4+x}}=(4+x)^{-1/2}$ , then factor: $=4^{-1/2}\big(1+\tfrac{x}{4}\big)^{-1/2}=\tfrac12\big(1+\tfrac{x}{4}\big)^{-1/2}$ .
Expand with $n=-\tfrac12$ , inside $\tfrac{x}{4}$ : $1-\tfrac12\tfrac{x}{4}+\tfrac{(-\frac12)(-\frac32)}{2}\big(\tfrac{x}{4}\big)^2=1-\tfrac{x}{8}+\tfrac{3x^2}{128}$ .
Multiply by $\tfrac12$ ; validity is $\big|\tfrac{x}{4}\big|<1$ , i.e. $|x|<4$ .

Answer

\tfrac12-\tfrac{x}{16}+\tfrac{3x^2}{256}+\cdots,\quad |x|<4

Worked example

First 3 terms of $(1+ax)^n$ are $1-24x+384x^2$ . Find $a$ and $n$

Match terms: the $x$ term gives $na=-24$ ; the $x^2$ term gives $\tfrac{n(n-1)}{2}a^2=384$ .
From the first, $a^2=\tfrac{576}{n^2}$ . Substitute: $\tfrac{n(n-1)}{2}\cdot\tfrac{576}{n^2}=384\Rightarrow 576(n-1)=768n\Rightarrow n=-3$ .
Then $a=\tfrac{-24}{n}=8$ .

Answer

n=-3,\quad a=8

Where the marks go

Common mistake

Don't forget to multiply the whole bracket expansion by

a^n

, and read the validity off the inside:

(a+x)^n

needs

|x|<|a|

, not

|x|<1

Common mistake

A reciprocal surd like

\tfrac1{\sqrt{4+x}}

is really a power:

(4+x)^{-1/2}

. Turn it into a single power before you do anything else. The binomial only works once it is written as one power.

Now you try

Expand $(2-3x)^{-2}$ in ascending powers of $x$ up to $x^2$ , simplify the coefficients. (9709/31 May/Jun 2020 Q2)

Your turn— tap to reveal the worked answer (9709/31 May/Jun 2020 Q2(a))

Pull the 2 out: $(2-3x)^{-2}=2^{-2}\big(1-\tfrac{3x}{2}\big)^{-2}=\tfrac14\big(1-\tfrac{3x}{2}\big)^{-2}$ . With $n=-2$ and inside $-\tfrac{3x}{2}$ :

$\tfrac14\Big[1+(-2)\big(-\tfrac{3x}{2}\big)+\tfrac{(-2)(-3)}{2}\big(-\tfrac{3x}{2}\big)^2\Big]=\tfrac14\big[1+3x+\tfrac{27}{4}x^2\big]$

Answer

\tfrac14+\tfrac{3x}{4}+\tfrac{27x^2}{16}+\cdots