AS & A Level · Practice questions · The binomial expansion

Choose the term.

Six original Cambridge-style questions on the binomial expansion of $(a + b)^n$ for positive integer $n$.

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

General term, then choose $r$.

Coefficients: $\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ or Pascal's triangle.
Powers: first term falls from $n$, second rises from 0; they add to $n$.
Brackets: $(-3x)^2 = 9x^2$; keep the sign and the number inside the power.
One term: simplify the power of $x$ in the general term, then solve for $r$.
Products: list every pair that makes the power you want, and add.

Recall

[2 marks]

(a) Find the value of $\binom{8}{3}$.

(b) Show that $\binom{n}{2} = \dfrac{n(n-1)}{2}$.

(a) $\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ✓
(b) $\dfrac{n!}{2!\,(n-2)!} = \dfrac{n(n-1)(n-2)!}{2 \times (n-2)!} = \dfrac{n(n-1)}{2}$ ✓

Calculation

[3 marks]

Expand $(2 - x)^4$ fully, simplifying each term.

Coefficients $1, 4, 6, 4, 1$. ✓
$2^4 + 4(2^3)(-x) + 6(2^2)(-x)^2 + 4(2)(-x)^3 + (-x)^4$ ✓
$= 16 - 32x + 24x^2 - 8x^3 + x^4$ ✓

Calculation

[3 marks]

Find the first three terms, in ascending powers of $x$, in the expansion of $(3 + 2x)^5$.

$3^5 = 243$ ✓
$\binom{5}{1}(3^4)(2x) = 5 \times 81 \times 2x = 810x$ ✓
$\binom{5}{2}(3^3)(2x)^2 = 10 \times 27 \times 4x^2 = 1080x^2$ ✓
$243 + 810x + 1080x^2$ ✓

Calculation

[3 marks]

Find the term independent of $x$ in the expansion of $\left(2x - \dfrac{1}{x^2}\right)^9$.

General term: $\binom{9}{r}(2x)^{9-r}\left(-\dfrac{1}{x^2}\right)^{r} = \binom{9}{r}2^{9-r}(-1)^{r}x^{9-3r}$ ✓
$9 - 3r = 0$, so $r = 3$. ✓
$\binom{9}{3} \times 2^6 \times (-1)^3 = 84 \times 64 \times (-1) = -5376$ ✓

Analysis

[4 marks]

(a) Find the first three terms, in ascending powers of $x$, of the expansion of $(2 - x)^6$.

(b) The coefficient of $x^2$ in the expansion of $(1 + kx)(2 - x)^6$ is zero. Find the value of the constant $k$.

(a) $64 + 6(32)(-x) + 15(16)x^2 = 64 - 192x + 240x^2$ ✓
(b) $x^2$ terms: $1 \times 240x^2$ and $kx \times (-192x)$. ✓
$240 - 192k = 0$, so $k = \dfrac{240}{192} = \dfrac{5}{4}$ ✓

Analysis

[5 marks]

A sum of $1000 is invested at 2% compound interest per year, so after 8 years it is worth $1000(1 + 0.02)^8$ dollars.

(a) Find the first three terms, in ascending powers of $x$, of $(1 + x)^8$.

(b) Use your answer to estimate the value of the investment after 8 years.

(a) $1 + 8x + 28x^2$ ✓
(b) $x = 0.02$: $1 + 0.16 + 28 \times 0.0004 = 1.1712$, so about $1171.20. ✓
(c) Calculator: $1000 \times 1.02^8 = 1171.66$. ✓
Error $= \dfrac{1171.66 - 1171.20}{1171.66} \times 100 = 0.0392\%$ (3 s.f.); the neglected terms are tiny because $x$ is small. ✓

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.

General term, then choose \(r\).