Home / A-Level Maths 9709 / Algebra / The binomial series for any rational n
Cambridge International AS & A Level · Topic 3.1
A-Level Maths 9709 / Topic 3.1 / AS & A Level

Infinite, but useful.

Paper 3 only. When \(n\) is negative or a fraction, \((1 + x)^n\) becomes an infinite series that is only valid for \(|x| < 1\). Expand it, adapt it to \((a + bx)^n\), state where it is valid, and use it to approximate.

Topic 3.1All algebra lessons →
The key idea

\((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{2!}x^2 + \dfrac{n(n - 1)(n - 2)}{3!}x^3 + \dots\), valid for \(|x| < 1\) when \(n\) is not a positive integer. For \((a + bx)^n\), first take out \(a^n\): \(a^n\left(1 + \dfrac{b}{a}x\right)^n\), valid for \(\left|\dfrac{bx}{a}\right| < 1\).

x y O 1 −1 (0, 1) valid: −1 < x < 1 y = (1 + x)⁻¹ y = 1 − x + x² − x³
Fig. 1: near x = 0 the first four terms of the series for (1 + x)^-1 (dashed) hug the curve; beyond x = 1 the series is not valid and the approximation falls away
Section 01

The series for any rational \(n\).

For a positive integer \(n\) the expansion stops after \(n + 1\) terms. For any other rational \(n\) the coefficients never reach zero, so the series goes on for ever, and it only adds up to \((1 + x)^n\) when \(|x| < 1\). You will normally be asked for the terms up to \(x^2\) or \(x^3\).

Worked example A negative index
Expand \((1 + x)^{-2}\) in ascending powers of \(x\), up to and including the term in \(x^3\).
  1. \(n = -2\): coefficient of \(x\) is \(-2\).
  2. Coefficient of \(x^2\): \(\dfrac{(-2)(-3)}{2} = 3\).
  3. Coefficient of \(x^3\): \(\dfrac{(-2)(-3)(-4)}{6} = -4\).
Answer: \(1 - 2x + 3x^2 - 4x^3\), valid for \(|x| < 1\)
Worked example A fractional index with a multiple of \(x\)
Expand \(\sqrt{1 - 4x}\) up to the term in \(x^3\), and state the set of values of \(x\) for which the expansion is valid.
  1. Write \((1 - 4x)^{1/2}\): \(n = \dfrac{1}{2}\) with \(-4x\) in place of \(x\).
  2. \(x\) term: \(\dfrac{1}{2}(-4x) = -2x\).
  3. \(x^2\) term: \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(-4x)^2 = -\dfrac{1}{8}(16x^2) = -2x^2\).
  4. \(x^3\) term: \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-4x)^3 = \dfrac{1}{16}(-64x^3) = -4x^3\).
  5. Valid when \(|-4x| < 1\), i.e. \(|x| < \dfrac{1}{4}\).
Answer: \(1 - 2x - 2x^2 - 4x^3\), valid for \(-\dfrac{1}{4} < x < \dfrac{1}{4}\)
Examiner trap

\((-4x)^2 = 16x^2\), not \(-4x^2\) or \(4x^2\). Write the bracket in every term, then square or cube both the number and the sign.

Section 02

Adapting to \((a + bx)^n\): take out \(a^n\).

The standard series needs a 1 in front. So factor out the constant first: \((a + bx)^n = a^n\left(1 + \dfrac{b}{a}x\right)^n\). Then expand with \(\dfrac{b}{a}x\) in place of \(x\), and multiply every term by \(a^n\).

Worked example Taking out a constant
Expand \((2 - x)^{-3}\) up to the term in \(x^3\), and state the range of validity.
  1. \((2 - x)^{-3} = 2^{-3}\left(1 - \dfrac{x}{2}\right)^{-3} = \dfrac{1}{8}\left(1 - \dfrac{x}{2}\right)^{-3}\).
  2. \(\left(1 - \dfrac{x}{2}\right)^{-3} = 1 + (-3)\left(-\dfrac{x}{2}\right) + \dfrac{(-3)(-4)}{2}\left(-\dfrac{x}{2}\right)^2 + \dfrac{(-3)(-4)(-5)}{6}\left(-\dfrac{x}{2}\right)^3\).
  3. \(= 1 + \dfrac{3x}{2} + 6 \cdot \dfrac{x^2}{4} - 10 \cdot \left(-\dfrac{x^3}{8}\right) = 1 + \dfrac{3}{2}x + \dfrac{3}{2}x^2 + \dfrac{5}{4}x^3\).
  4. Multiply by \(\dfrac{1}{8}\).
  5. Valid when \(\left|\dfrac{x}{2}\right| < 1\), i.e. \(|x| < 2\).
Answer: \(\dfrac{1}{8} + \dfrac{3}{16}x + \dfrac{3}{16}x^2 + \dfrac{5}{32}x^3\), valid for \(|x| < 2\)
Worked example Approximating a surd
Use the expansion \(\sqrt{1 - 4x} \approx 1 - 2x - 2x^2 - 4x^3\) with \(x = 0.01\) to find an approximation for \(\sqrt{6}\).
  1. With \(x = 0.01\): \(\sqrt{1 - 0.04} = \sqrt{0.96}\), and \(x = 0.01\) is inside \(|x| < \dfrac{1}{4}\).
  2. \(\sqrt{0.96} = \sqrt{\dfrac{96}{100}} = \dfrac{\sqrt{16 \times 6}}{10} = \dfrac{4\sqrt{6}}{10}\).
  3. Series: \(1 - 0.02 - 0.0002 - 0.000\,004 = 0.979\,796\).
  4. So \(\sqrt{6} \approx \dfrac{10}{4} \times 0.979\,796 = 2.449\,49\).
Answer: \(\sqrt{6} \approx 2.449\,49\) (the true value is \(2.449\,489\,7\dots\))
Your turn
Expand \(\dfrac{1}{\sqrt{4 + x}}\) up to the term in \(x^2\), and state the values of \(x\) for which it is valid.
Examiner trap

\((2 - x)^{-3}\) is NOT \(2\left(1 - \dfrac{x}{2}\right)^{-3}\). The factor taken out is \(2^{-3} = \dfrac{1}{8}\): it carries the same power as the bracket.

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What is the coefficient of \(x^2\) in the expansion of \((1 + x)^{-3}\)?

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For which values of \(x\) is the expansion of \((1 + 5x)^{1/3}\) valid?

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The first two terms of the expansion of \((1 - 2x)^{-1/2}\) are

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To expand \((9 + x)^{1/2}\), the first step is to write it as

Section 03

With partial fractions.

A rational function is expanded by splitting it into partial fractions, expanding each piece, and adding. The expansion is valid only where EVERY piece is valid, so take the narrower range.

Worked example Split, expand, add
Given \(\dfrac{8x - 1}{(x + 1)(2x - 1)} \equiv \dfrac{3}{1 + x} + \dfrac{2}{2x - 1}\), expand the function in ascending powers of \(x\) up to \(x^2\) and state the range of validity.
  1. \(\dfrac{3}{1 + x} = 3(1 + x)^{-1} = 3(1 - x + x^2) = 3 - 3x + 3x^2\), valid for \(|x| < 1\).
  2. \(\dfrac{2}{2x - 1} = -\dfrac{2}{1 - 2x} = -2(1 - 2x)^{-1} = -2(1 + 2x + 4x^2) = -2 - 4x - 8x^2\), valid for \(|x| < \dfrac{1}{2}\).
  3. Add: \((3 - 2) + (-3 - 4)x + (3 - 8)x^2\).
  4. Both pieces are valid only when \(|x| < \dfrac{1}{2}\).
Answer: \(1 - 7x - 5x^2\), valid for \(|x| < \dfrac{1}{2}\)
Worked example Finding an unknown from a coefficient
In the expansion of \((1 + ax)^{-1/2}\), the coefficient of \(x^2\) is \(\dfrac{27}{2}\). Find the possible values of \(a\).
  1. Coefficient of \(x^2\): \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}a^2 = \dfrac{3}{8}a^2\).
  2. \(\dfrac{3}{8}a^2 = \dfrac{27}{2}\), so \(a^2 = 36\).
Answer: \(a = 6\) or \(a = -6\)
Examiner trap

Write \(\dfrac{2}{2x - 1}\) as \(-2(1 - 2x)^{-1}\) before expanding: the standard series needs the constant term in the bracket to be \(+1\). Expanding \((2x - 1)^{-1}\) 'as if' it were \((1 + 2x)^{-1}\) gives the wrong signs.

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Find the first three terms in the expansion of \((1 + 3x)^{1/3}\) in ascending powers of \(x\).

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Expand \((3 + 2x)^{-2}\) up to the term in \(x^2\).

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\(\dfrac{1 + x}{(1 - 2x)^2}\) is expanded in ascending powers of \(x\). Find the coefficient of \(x^2\).

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\(f(x) = \dfrac{3}{1 + x} + \dfrac{2}{2x - 1}\) is expanded in ascending powers of \(x\). For which set of values of \(x\) is the expansion valid?

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Binomial series, expanded.

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Six original Cambridge-style questions
Expanding \((1 + x)^n\) for negative and fractional \(n\), adapting to \((a + bx)^n\), validity, approximation and partial fractions.