AS & A Level · Practice questions · The binomial series for any rational n

Terms that never stop.

Paper 3 only. Six original Cambridge-style questions on the binomial series, ending with a partial fractions and expansion problem.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Make it \((1 + \dots)^n\), then expand.

Standard form: \((a + bx)^n = a^n\left(1 + \dfrac{b}{a}x\right)^n\).
Coefficients: \(1,\ n,\ \dfrac{n(n - 1)}{2!},\ \dfrac{n(n - 1)(n - 2)}{3!}\).
Brackets: raise the whole term, sign and number, to each power.
Validity: \(\left|\dfrac{bx}{a}\right| < 1\); for sums take the narrowest range.
Products: collect every pair of terms that multiply to the power you need.

Recall

[3 marks]

Expand \((1 - 2x)^{-1/2}\) in ascending powers of \(x\), up to and including the term in \(x^3\).

\(x\): \(\left(-\dfrac{1}{2}\right)(-2x) = x\). ✓
\(x^2\): \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(4x^2) = \dfrac{3}{8} \cdot 4x^2 = \dfrac{3}{2}x^2\). ✓
\(x^3\): \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(-8x^3) = -\dfrac{5}{16} \cdot (-8x^3) = \dfrac{5}{2}x^3\). ✓
\(1 + x + \dfrac{3}{2}x^2 + \dfrac{5}{2}x^3\), valid for \(|x| < \dfrac{1}{2}\). ✓

Calculation

[4 marks]

Expand \(\sqrt[3]{8 + 3x}\) up to the term in \(x^2\), and state the set of values of \(x\) for which the expansion is valid.

\((8 + 3x)^{1/3} = 8^{1/3}\left(1 + \dfrac{3x}{8}\right)^{1/3} = 2\left(1 + \dfrac{3x}{8}\right)^{1/3}\). ✓
\(= 2\left(1 + \dfrac{1}{3} \cdot \dfrac{3x}{8} + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2} \cdot \dfrac{9x^2}{64}\right) = 2\left(1 + \dfrac{x}{8} - \dfrac{x^2}{64}\right)\). ✓
\(= 2 + \dfrac{x}{4} - \dfrac{x^2}{32}\). ✓
Valid for \(\left|\dfrac{3x}{8}\right| < 1\), i.e. \(|x| < \dfrac{8}{3}\). ✓

Calculation

[3 marks]

Find the coefficient of \(x^3\) in the expansion of \((1 + 3x)^{1/3}\).

Coefficient: \(\dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{6} \times 3^3 = \dfrac{10}{27 \times 6} \times 27\). ✓
\(= \dfrac{10}{6} = \dfrac{5}{3}\). ✓

Analysis

[4 marks]

(a) Expand \((1 + x)^{1/2}\) up to the term in \(x^2\).

(b) By substituting \(x = \dfrac{1}{16}\), find an approximation to \(\sqrt{17}\), giving your answer to 4 decimal places.

(a) \(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2\), valid for \(|x| < 1\). ✓
(b) \(\sqrt{1 + \dfrac{1}{16}} = \sqrt{\dfrac{17}{16}} = \dfrac{\sqrt{17}}{4}\). ✓
Series: \(1 + \dfrac{1}{32} - \dfrac{1}{2048} = 1.030\,761\dots\) ✓
\(\sqrt{17} \approx 4 \times 1.030\,761 = 4.1230\) (4 d.p.). The true value is \(4.1231\dots\) ✓

Analysis

[4 marks]

In the expansion of \((1 + ax)^{n}\), the coefficient of \(x\) is \(-6\) and the coefficient of \(x^2\) is \(27\). Find \(n\) and \(a\).

\(na = -6\) and \(\dfrac{n(n - 1)}{2}a^2 = 27\). ✓
Write the second as \(\dfrac{(na)^2}{2} \cdot \dfrac{n - 1}{n} = 27\): \(18 \cdot \dfrac{n - 1}{n} = 27\). ✓
\(18n - 18 = 27n\), so \(n = -2\). ✓
\(a = \dfrac{-6}{n} = 3\). (Check: \((1 + 3x)^{-2} = 1 - 6x + 27x^2 - \dots\)) ✓

Analysis

[7 marks]

Let \(f(x) = \dfrac{5x^2 + 3x + 7}{(x + 2)(x^2 + 3)}\).

(a) Given that \(f(x) \equiv \dfrac{3}{x + 2} + \dfrac{2x - 1}{x^2 + 3}\), find the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).

(b) State the set of values of \(x\) for which the expansion is valid.

(a) \(\dfrac{3}{x + 2} = \dfrac{3}{2}\left(1 + \dfrac{x}{2}\right)^{-1} = \dfrac{3}{2}\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4}\right) = \dfrac{3}{2} - \dfrac{3}{4}x + \dfrac{3}{8}x^2\). ✓
\(\dfrac{1}{x^2 + 3} = \dfrac{1}{3}\left(1 + \dfrac{x^2}{3}\right)^{-1} = \dfrac{1}{3} - \dfrac{x^2}{9} + \dots\) ✓
\((2x - 1)\left(\dfrac{1}{3} - \dfrac{x^2}{9}\right) = -\dfrac{1}{3} + \dfrac{2}{3}x + \dfrac{1}{9}x^2 + \dots\) (terms up to \(x^2\)). ✓
Add: \(\left(\dfrac{3}{2} - \dfrac{1}{3}\right) + \left(-\dfrac{3}{4} + \dfrac{2}{3}\right)x + \left(\dfrac{3}{8} + \dfrac{1}{9}\right)x^2 = \dfrac{7}{6} - \dfrac{1}{12}x + \dfrac{35}{72}x^2\). ✓
(b) First piece: \(|x| < 2\). Second: \(\dfrac{x^2}{3} < 1\), i.e. \(|x| < \sqrt{3}\). Both: \(|x| < \sqrt{3}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.