IGCSE · Practice questions · The binomial expansion and its general term

Write the general term, then solve for r.

Six original Cambridge-style questions on the binomial expansion of \((a + b)^{n}\): Pascal's triangle and \({}^{n}C_{r}\) coefficients, the general term \({}^{n}C_{r}\,a^{n-r}b^{r}\), constant terms, unknown constants, products of brackets and a simple approximation, all without a calculator.

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

The method

Brackets first, then the general term.

Identify \(a\), \(b\) and \(n\); write \(b\) with its sign and coefficient in brackets, such as \(b = \left(-\dfrac{x}{2}\right)\).
Full expansion: coefficients from Pascal's triangle or \({}^{n}C_{r}\); powers of \(a\) fall as powers of \(b\) rise.
One term: write \({}^{n}C_{r}\,a^{n-r}b^{r}\), collect the power of \(x\) as an expression in \(r\), set it equal to the power wanted (0 for the constant term) and solve.
Unknowns: set the coefficient equal to the given value and solve; watch for \(\pm\) when you take a square root.
Products: list every pairing of terms that makes the power wanted, then add the results.

Recall

[3 marks]

Expand \((1 + 2x)^{4}\) fully, simplifying each coefficient.

Row 4 of Pascal's triangle: 1, 4, 6, 4, 1, with \(a = 1\) and \(b = (2x)\). ✓
\(1 + 4(2x) + 6(2x)^{2} + 4(2x)^{3} + (2x)^{4}\) ✓
\(= 1 + 8x + 24x^{2} + 32x^{3} + 16x^{4}\). ✓

Calculation

[3 marks]

Find the first three terms, in ascending powers of \(x\), in the expansion of \(\left(3 - \dfrac{x}{3}\right)^{6}\). Give each coefficient as an integer.

Constant term: \(3^{6} = 729\). ✓
\(x\) term: \({}^{6}C_{1} \times 3^{5} \times \left(-\dfrac{x}{3}\right) = 6 \times 243 \times \left(-\dfrac{x}{3}\right) = -486x\). ✓
\(x^{2}\) term: \({}^{6}C_{2} \times 3^{4} \times \dfrac{x^{2}}{9} = 15 \times 81 \times \dfrac{x^{2}}{9} = 135x^{2}\). ✓
First three terms: \(729 - 486x + 135x^{2}\). ✓

Calculation

[3 marks]

Find the coefficient of \(x^{5}\) in the expansion of \((2x + 3)^{7}\).

General term: \({}^{7}C_{r}\,(2x)^{7-r}\,3^{r}\). The power of \(x\) is \(7 - r = 5\), so \(r = 2\). ✓
\({}^{7}C_{2} = 21\), \(2^{5} = 32\), \(3^{2} = 9\). ✓
Coefficient: \(21 \times 32 \times 9 = 6048\). ✓

Calculation

[4 marks]

Find the term independent of \(x\) in the expansion of \(\left(x^{2} + \dfrac{3}{x}\right)^{6}\).

General term: \({}^{6}C_{r}\,(x^{2})^{6-r}\left(\dfrac{3}{x}\right)^{r} = {}^{6}C_{r}\,3^{r}\,x^{12-2r}\,x^{-r}\). ✓
Power of \(x\): \(12 - 3r\). Independent of \(x\): \(12 - 3r = 0\), so \(r = 4\). ✓
\({}^{6}C_{4} = {}^{6}C_{2} = 15\) and \(3^{4} = 81\). ✓
Term: \(15 \times 81 = 1215\). ✓

Analysis

[5 marks]

In the expansion of \((2 + kx)^{5}\), where \(k\) is a non-zero constant, the coefficient of \(x^{2}\) is equal to the coefficient of \(x^{3}\).

(a) Show that \(k = 2\).

(b) Hence find the coefficient of \(x^{2}\) in the expansion of \((1 - x)(2 + kx)^{5}\).

(a) Coefficient of \(x^{2}\): \({}^{5}C_{2} \times 2^{3} \times k^{2} = 10 \times 8k^{2} = 80k^{2}\). ✓
Coefficient of \(x^{3}\): \({}^{5}C_{3} \times 2^{2} \times k^{3} = 10 \times 4k^{3} = 40k^{3}\). ✓
\(80k^{2} = 40k^{3}\); \(k \ne 0\), so divide by \(40k^{2}\): \(k = 2\). ✓
(b) \((2 + 2x)^{5} = 32 + 5 \times 16 \times 2x + 10 \times 8 \times 4x^{2} + \cdots = 32 + 160x + 320x^{2} + \cdots\). ✓
\(x^{2}\) terms of the product: \(1 \times 320x^{2}\) and \((-x) \times 160x\), so the coefficient is \(320 - 160 = 160\). ✓

Analysis

[6 marks]

(a) Find the first three terms, in ascending powers of \(x\), in the expansion of \(\left(1 - \dfrac{x}{2}\right)^{8}\).

(b) Hence find the coefficient of \(x^{2}\) in the expansion of \((3 + 2x)\left(1 - \dfrac{x}{2}\right)^{8}\).

(c) Use your answer to (a), with a suitable value of \(x\), to estimate \(0.995^{8}\), giving your answer to 4 decimal places.

(a) \(1 + 8\left(-\dfrac{x}{2}\right) + 28\left(-\dfrac{x}{2}\right)^{2} = 1 - 4x + 28 \times \dfrac{x^{2}}{4} = 1 - 4x + 7x^{2}\). ✓
(b) \(x^{2}\) terms: \(3 \times 7x^{2}\) and \(2x \times (-4x)\), so the coefficient is \(21 - 8 = 13\). ✓
(c) \(1 - \dfrac{x}{2} = 0.995\) gives \(\dfrac{x}{2} = 0.005\), so \(x = 0.01\). ✓
\(1 - 4(0.01) + 7(0.01)^{2} = 1 - 0.04 + 0.0007 = 0.9607\). ✓
So \(0.995^{8} \approx 0.9607\) (the true value is \(0.96069\dots\), so the first three terms are enough here). ✓

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.