Binomial Expansion for Rational Powers

A2 · P3 · 11 min

In P1 you expanded $(1 + x)^n$ for a positive whole number $n$ , and the expansion stopped after $n + 1$ terms. In P3, $n$ can be any rational number: negative, like $(1 + 2x)^{-3}$ , or fractional, like $\sqrt{1 - 4x} = (1 - 4x)^{1/2}$ . The expansion then never stops. It becomes an infinite series, which is only valid for small enough $x$ . P3 questions ask for the first three or four terms, the set of values of $x$ for which the expansion is valid, and often an approximation or an unknown constant.

A series that never ends

You have already met one example. The sum to infinity of a geometric progression (see geometric progressions) with first term $1$ and common ratio $-x$ is

1 - x + x^2 - x^3 + \cdots = \frac{1}{1 + x} = (1 + x)^{-1}, \qquad \text{valid for } |x| < 1.

So $(1 + x)^{-1}$ "expands" to an infinite series, but only when $|x| < 1$ . For $x = 2$ the left side is $1 - 2 + 4 - 8 + \cdots$ , which does not settle to anything, while the right side is $\tfrac{1}{3}$ .

The graph shows this. Near $x = 0$ the first four terms $1 - x + x^2 - x^3$ hug the curve $y = \dfrac{1}{1 + x}$ closely. Towards $x = \pm 1$ they peel away, and outside that interval they are useless.

y = 1/(1 + x) y = 1 - x + x^2 - x^3 (-1, -1) -- (-1, 5) (1, -1) -- (1, 5)

The general binomial series

The P1 formula still works when $n$ is not a positive integer. The coefficients are written out in full, because the $\dbinom{n}{r}$ notation is only defined for whole numbers.

Key result

For any rational number $n$ , and $|x| < 1$ ,

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

This formula is given in the list of formulae (MF19), but you must know how to use it.

If $n$ is a positive integer the series stops (eventually a factor $(n - n)$ appears). Otherwise it is infinite and valid only for $|x| < 1$ .

Why does it never stop? Each coefficient contains the factors $n, n - 1, n - 2, \ldots$ . For a whole number $n$ one of them is eventually $0$ , killing every later term. For $n = -3$ or $n = \tfrac{1}{2}$ , none of them is ever zero.

Tip

The general term is not required in P3. You only ever need the first few terms, usually up to $x^2$ or $x^3$ .

Expanding $(1 + kx)^n$

Replace $x$ by $kx$ throughout. The whole of $kx$ is raised to each power: $(kx)^2 = k^2x^2$ , $(kx)^3 = k^3x^3$ . Brackets are essential here.

Expanding (1 + kx) to the power n

Identify $n$ and the " $x$ " of the formula, here $u = kx$ (including its sign).
Write the terms with brackets: $1 + n(u) + \dfrac{n(n - 1)}{2}(u)^2 + \dfrac{n(n - 1)(n - 2)}{6}(u)^3$ .
Work out each coefficient as a single simplified fraction.
State the validity: $|u| < 1$ , i.e. $|kx| < 1$ , i.e. $|x| < \dfrac{1}{|k|}$ .

A negative power

Expand $(1 + 2x)^{-3}$ in ascending powers of $x$ , up to and including the term in $x^3$ , and state the set of values of $x$ for which the expansion is valid.

Solution

$n = -3$ , $u = 2x$ .

(1 + 2x)^{-3} = 1 + (-3)(2x) + \frac{(-3)(-4)}{2}(2x)^2 + \frac{(-3)(-4)(-5)}{6}(2x)^3 + \cdots

Term by term: $(-3)(2x) = -6x$ ; $\ 6 \times 4x^2 = 24x^2$ ; $\ (-10) \times 8x^3 = -80x^3$ .

(1 + 2x)^{-3} = 1 - 6x + 24x^2 - 80x^3 + \cdots

Valid for $|2x| < 1$ , that is $|x| < \tfrac{1}{2}$ , or $-\tfrac{1}{2} < x < \tfrac{1}{2}$ .

A fractional power

Expand $\sqrt{1 - 4x}$ up to and including the term in $x^3$ , simplifying the coefficients, and state the range of validity.

Solution

$\sqrt{1 - 4x} = (1 - 4x)^{1/2}$ , so $n = \tfrac{1}{2}$ , $u = -4x$ .

1 + \tfrac{1}{2}(-4x) + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}(-4x)^2 + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{6}(-4x)^3

The coefficients: $\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2} = -\tfrac{1}{8}$ and $\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{6} = \dfrac{3/8}{6} = \tfrac{1}{16}$ .

So the terms are $-2x$ , $\ -\tfrac{1}{8} \times 16x^2 = -2x^2$ , and $\tfrac{1}{16} \times (-64x^3) = -4x^3$ :

\sqrt{1 - 4x} = 1 - 2x - 2x^2 - 4x^3 + \cdots

Valid for $|-4x| < 1$ , i.e. $|x| < \tfrac{1}{4}$ .

Watch out

$(-4x)^2 = +16x^2$ and $(-4x)^3 = -64x^3$ . Losing the sign of $u$ inside the powers is the most common error in this topic. Keep $u$ in brackets until the last line.

Expanding $(a + bx)^n$

The formula needs a $1$ in front. So take out the factor $a$ first, and remember that it comes out raised to the power $n$ :

Key result

(a + bx)^n = a^n\left(1 + \frac{b}{a}x\right)^n, \qquad \text{valid for } \left|\frac{bx}{a}\right| < 1, \ \text{ i.e. } |x| < \left|\frac{a}{b}\right|.

For example, $\sqrt{4 - x} = 4^{1/2}\left(1 - \tfrac{x}{4}\right)^{1/2} = 2\left(1 - \tfrac{x}{4}\right)^{1/2}$ , and $(2 + 3x)^{-2} = 2^{-2}\left(1 + \tfrac{3x}{2}\right)^{-2} = \tfrac{1}{4}\left(1 + \tfrac{3x}{2}\right)^{-2}$ .

Taking out a factor

Expand $\dfrac{1}{(2 + 3x)^2}$ in ascending powers of $x$ up to the term in $x^3$ , and state the set of values of $x$ for which the expansion is valid.

Solution

(2 + 3x)^{-2} = 2^{-2}\left(1 + \tfrac{3}{2}x\right)^{-2} = \tfrac{1}{4}\left(1 + \tfrac{3}{2}x\right)^{-2}.

With $n = -2$ and $u = \tfrac{3}{2}x$ :

\left(1 + \tfrac{3}{2}x\right)^{-2} = 1 + (-2)\left(\tfrac{3}{2}x\right) + \frac{(-2)(-3)}{2}\left(\tfrac{3}{2}x\right)^2 + \frac{(-2)(-3)(-4)}{6}\left(\tfrac{3}{2}x\right)^3 + \cdots

= 1 - 3x + 3 \cdot \tfrac{9}{4}x^2 - 4 \cdot \tfrac{27}{8}x^3 + \cdots = 1 - 3x + \tfrac{27}{4}x^2 - \tfrac{27}{2}x^3 + \cdots

Multiply by $\tfrac{1}{4}$ :

\frac{1}{(2 + 3x)^2} = \frac{1}{4} - \frac{3}{4}x + \frac{27}{16}x^2 - \frac{27}{8}x^3 + \cdots

Valid for $\left|\tfrac{3}{2}x\right| < 1$ , i.e. $|x| < \tfrac{2}{3}$ .

Watch out

Two classic errors with $(a + bx)^n$ :

Taking out $a$ instead of $a^n$ . $(2 + 3x)^{-2}$ is $\tfrac{1}{4}(\ldots)$ , not $2(\ldots)$ or $\tfrac{1}{2}(\ldots)$ .
Stating the validity as $|x| < 1$ . The condition is on $u$ , so here it is $\left|\tfrac{3x}{2}\right| < 1$ , i.e. $|x| < \tfrac{2}{3}$ .

Adapting the series: powers of $x^2$

Sometimes $u$ is a multiple of $x^2$ . Nothing changes except that the powers jump in twos: $u^2$ is an $x^4$ term.

An expansion in powers of x squared

Expand $\dfrac{1}{\sqrt{4 - x^2}}$ in ascending powers of $x$ up to and including the term in $x^4$ , and state the set of values of $x$ for which the expansion is valid.

Solution

(4 - x^2)^{-1/2} = 4^{-1/2}\left(1 - \tfrac{x^2}{4}\right)^{-1/2} = \tfrac{1}{2}\left(1 - \tfrac{x^2}{4}\right)^{-1/2}.

With $n = -\tfrac{1}{2}$ and $u = -\tfrac{x^2}{4}$ :

\left(1 - \tfrac{x^2}{4}\right)^{-1/2} = 1 + \left(-\tfrac{1}{2}\right)\left(-\tfrac{x^2}{4}\right) + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2}\left(-\tfrac{x^2}{4}\right)^2 + \cdots = 1 + \tfrac{1}{8}x^2 + \tfrac{3}{8} \cdot \tfrac{x^4}{16} + \cdots

\frac{1}{\sqrt{4 - x^2}} = \frac{1}{2} + \frac{1}{16}x^2 + \frac{3}{256}x^4 + \cdots

Valid for $\left|\tfrac{x^2}{4}\right| < 1$ , i.e. $x^2 < 4$ , i.e. $|x| < 2$ .

Finding unknown constants

If the expansion is given, comparing coefficients produces equations for unknown values of $n$ or $k$ .

Finding n and a

The first three terms in the expansion of $(1 + ax)^n$ , in ascending powers of $x$ , are $1 - 6x + 27x^2$ . Find the values of $a$ and $n$ , and state the set of values of $x$ for which the expansion is valid.

Solution

Compare coefficients with $1 + n(ax) + \dfrac{n(n - 1)}{2}(ax)^2$ :

na = -6 \qquad\qquad \frac{n(n - 1)}{2}a^2 = 27.

From the first, $a = -\dfrac{6}{n}$ , so $a^2 = \dfrac{36}{n^2}$ . Substitute:

\frac{n(n - 1)}{2} \cdot \frac{36}{n^2} = 27 \quad\Rightarrow\quad \frac{18(n - 1)}{n} = 27 \quad\Rightarrow\quad 18n - 18 = 27n \quad\Rightarrow\quad n = -2.

Then $a = -\dfrac{6}{-2} = 3$ . So the expression is $(1 + 3x)^{-2}$ .

Check the $x^2$ term: $\dfrac{(-2)(-3)}{2}(3x)^2 = 3 \times 9x^2 = 27x^2$ .

Valid for $|3x| < 1$ , i.e. $|x| < \tfrac{1}{3}$ .

Approximations

When $x$ is small, the terms $x^2, x^3, \ldots$ shrink fast, so a few terms give an excellent approximation. To approximate a specific number, choose $x$ so that the expression becomes that number, check that $x$ is inside the range of validity, and substitute.

Approximating a square root

Use the expansion $\sqrt{1 - 4x} = 1 - 2x - 2x^2 - 4x^3 + \cdots$ with $x = 0.01$ to find an approximation to $\sqrt{6}$ , giving your answer to 5 significant figures.

Solution

$x = 0.01$ lies inside $|x| < \tfrac{1}{4}$ , so the expansion is valid.

\sqrt{1 - 0.04} = \sqrt{0.96} \approx 1 - 0.02 - 0.0002 - 0.000004 = 0.979796.

Now relate $\sqrt{0.96}$ to $\sqrt{6}$ : $0.96 = \dfrac{96}{100} = \dfrac{16 \times 6}{100}$ , so $\sqrt{0.96} = \dfrac{4\sqrt{6}}{10} = \dfrac{2\sqrt{6}}{5}$ . Therefore

\sqrt{6} = \tfrac{5}{2}\sqrt{0.96} \approx \tfrac{5}{2} \times 0.979796 = 2.44949 = 2.4495 \ \text{(5 s.f.)}.

(The true value is $2.449489\ldots$ .)

Watch out

An expansion is meaningless outside its range of validity. The expansion of $\sqrt{1 - 4x}$ is valid only for $|x| < \tfrac{1}{4}$ . Putting $x = 0.5$ into its first four terms gives $1 - 1 - 0.5 - 0.5 = -1$ , yet $\sqrt{1 - 2}$ is not even a real number. Always check that the value you substitute is inside the range of validity.

Common mistakes

Watch out

Dividing by the wrong factorial. The $x^3$ coefficient is divided by $3! = 6$ , not $3$ . The $x^2$ coefficient is divided by $2! = 2$ .

Watch out

Dropping the minus signs in $n(n - 1)(n - 2)$ . For $n = -\tfrac{1}{2}$ the factors are $-\tfrac{1}{2}, -\tfrac{3}{2}, -\tfrac{5}{2}$ , all negative. For $n = \tfrac{1}{2}$ they are $\tfrac{1}{2}, -\tfrac{1}{2}, -\tfrac{3}{2}$ . Write each factor out.

Watch out

Writing roots and reciprocals as powers incorrectly. $\dfrac{1}{\sqrt{1 + x}} = (1 + x)^{-1/2}$ ; $\sqrt[3]{(1 + x)^2} = (1 + x)^{2/3}$ ; $\dfrac{1}{(2 - x)^3} = (2 - x)^{-3}$ .

Exam tip

"In ascending powers of $x$ " means constant first, then $x$ , $x^2$ , $x^3$ .
"Up to and including the term in $x^3$ " means four terms (if none are zero). Extra correct terms are ignored, but a wrong extra term can cost the accuracy mark.
Coefficients must be simplified exact fractions such as $\tfrac{27}{16}$ , not decimals.
Typical marks: one for the correct unsimplified $x$ and $x^2$ terms, one for the $x^3$ term, one for full simplification, and one for taking out $a^n$ correctly when needed.
When asked "state the set of values of $x$ for which the expansion is valid", give $|x| < \tfrac{2}{3}$ or $-\tfrac{2}{3} < x < \tfrac{2}{3}$ .

Summary

$(1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{2!}x^2 + \dfrac{n(n - 1)(n - 2)}{3!}x^3 + \cdots$ for rational $n$ and $|x| < 1$ .
For non-integer or negative $n$ the series is infinite and only valid for $|x| < 1$ .
Replace $x$ by $u = kx$ (or $kx^2$ ) with brackets: $(kx)^r = k^rx^r$ .
$(a + bx)^n = a^n\left(1 + \tfrac{b}{a}x\right)^n$ , valid for $|x| < \left|\tfrac{a}{b}\right|$ .
Given terms of an expansion, compare coefficients to find unknowns.
For approximations, choose $x$ inside the range of validity.

Practice

Question

Expand $(1 - 3x)^{-2}$ in ascending powers of $x$ up to the term in $x^3$ , and state the set of values of $x$ for which the expansion is valid.
Expand $\sqrt[3]{1 + 6x}$ up to the term in $x^2$ , and state the validity.
Expand $\dfrac{1}{\sqrt{9 + x}}$ up to the term in $x^2$ , and state the validity.
Expand $\dfrac{1}{(2 - x)^3}$ up to the term in $x^2$ , and state the validity.
Expand $\sqrt{1 + 2x^2}$ up to the term in $x^4$ , and state the set of values of $x$ for which the expansion is valid.
The coefficient of $x^2$ in the expansion of $(1 + ax)^{-1/2}$ is $6$ , where $a > 0$ . Find $a$ and the coefficient of $x^3$ .
The first three terms in the expansion of $(1 + ax)^n$ are $1 + 2x + 6x^2$ . Find $a$ and $n$ , state the validity, and find the coefficient of $x^3$ .
Expand $(1 - 2x)^{-1/2}$ up to the term in $x^3$ . By substituting $x = 0.02$ , find an approximation to $\sqrt{6}$ , giving your answer to 4 decimal places.

Answers

$n = -2$ , $u = -3x$ : $1 + (-2)(-3x) + \dfrac{(-2)(-3)}{2}(9x^2) + \dfrac{(-2)(-3)(-4)}{6}(-27x^3) = 1 + 6x + 27x^2 + 108x^3$ . Valid for $|x| < \tfrac{1}{3}$ .
$n = \tfrac{1}{3}$ , $u = 6x$ : $1 + \tfrac{1}{3}(6x) + \dfrac{\tfrac{1}{3}\left(-\tfrac{2}{3}\right)}{2}(36x^2) = 1 + 2x - \tfrac{1}{9}(36x^2) = 1 + 2x - 4x^2$ . Valid for $|x| < \tfrac{1}{6}$ .
$(9 + x)^{-1/2} = \tfrac{1}{3}\left(1 + \tfrac{x}{9}\right)^{-1/2} = \tfrac{1}{3}\left[1 - \tfrac{1}{2} \cdot \tfrac{x}{9} + \tfrac{3}{8} \cdot \tfrac{x^2}{81}\right] = \tfrac{1}{3} - \tfrac{x}{54} + \tfrac{x^2}{648}$ . Valid for $|x| < 9$ .
$(2 - x)^{-3} = \tfrac{1}{8}\left(1 - \tfrac{x}{2}\right)^{-3} = \tfrac{1}{8}\left[1 + (-3)\left(-\tfrac{x}{2}\right) + 6 \cdot \tfrac{x^2}{4}\right] = \tfrac{1}{8} + \tfrac{3}{16}x + \tfrac{3}{16}x^2$ . Valid for $|x| < 2$ .
$n = \tfrac{1}{2}$ , $u = 2x^2$ : $1 + \tfrac{1}{2}(2x^2) - \tfrac{1}{8}(4x^4) = 1 + x^2 - \tfrac{1}{2}x^4$ . Valid for $2x^2 < 1$ , i.e. $|x| < \tfrac{1}{\sqrt{2}}$ .
$x^2$ coefficient: $\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2}a^2 = \tfrac{3}{8}a^2 = 6$ , so $a^2 = 16$ and $a = 4$ . $x^3$ coefficient: $\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)\left(-\tfrac{5}{2}\right)}{6}a^3 = -\tfrac{5}{16} \times 64 = -20$ .
$na = 2$ and $\tfrac{n(n - 1)}{2}a^2 = 6$ . With $a^2 = \tfrac{4}{n^2}$ : $\tfrac{2(n - 1)}{n} = 6$ , so $2n - 2 = 6n$ , $n = -\tfrac{1}{2}$ and $a = -4$ . The expression is $(1 - 4x)^{-1/2}$ , valid for $|x| < \tfrac{1}{4}$ . $x^3$ coefficient: $-\tfrac{5}{16} \times (-4)^3 = -\tfrac{5}{16} \times (-64) = 20$ .
$n = -\tfrac{1}{2}$ , $u = -2x$ : $1 + x + \tfrac{3}{8}(4x^2) - \tfrac{5}{16}(-8x^3) = 1 + x + \tfrac{3}{2}x^2 + \tfrac{5}{2}x^3$ , valid for $|x| < \tfrac{1}{2}$ . At $x = 0.02$ : $(0.96)^{-1/2} \approx 1 + 0.02 + 0.0006 + 0.00002 = 1.02062$ . Since $\sqrt{0.96} = \tfrac{2\sqrt{6}}{5}$ , $(0.96)^{-1/2} = \tfrac{5}{2\sqrt{6}} = \tfrac{5\sqrt{6}}{12}$ . So $\sqrt{6} \approx \tfrac{12}{5} \times 1.02062 = 2.44949$ , i.e. $2.4495$ (4 d.p.).

A series that never ends

The general binomial series

Expanding (1+kx)n(1 + kx)^n(1+kx)n

Expanding (a+bx)n(a + bx)^n(a+bx)n

Adapting the series: powers of x2x^2x2

Finding unknown constants

Approximations

Common mistakes

Summary

Practice

Console

Expanding $(1 + kx)^n$

Expanding $(a + bx)^n$

Adapting the series: powers of $x^2$