| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expand and state validity |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial expansion formula for negative indices, requiring substitution into the standard formula and algebraic simplification. While it involves multiple steps and careful arithmetic with fractions, it's a routine C4 exercise with no problem-solving or novel insight required. The validity part is standard recall. Slightly easier than average due to being purely procedural. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^{-3}(1 - \frac{3}{2}x)^{-3} = \frac{1}{8}(1 - \frac{3}{2}x)^{-3}\) | B1 | |
| \(= \frac{1}{8}[1 + (-3)(-\frac{3}{2}x) + \frac{(-3)(-4)}{2}(-\frac{3}{2}x)^2 + \frac{(-3)(-4)(-5)}{3 \times 2}(-\frac{3}{2}x)^3 + \ldots]\) | M1 | |
| \(= \frac{1}{8} + \frac{9}{16}x + \frac{27}{16}x^2 + \frac{135}{32}x^3 + \ldots\) | A3 | |
| \(\ | x\ | < \frac{2}{3}\) |
$2^{-3}(1 - \frac{3}{2}x)^{-3} = \frac{1}{8}(1 - \frac{3}{2}x)^{-3}$ | B1 |
$= \frac{1}{8}[1 + (-3)(-\frac{3}{2}x) + \frac{(-3)(-4)}{2}(-\frac{3}{2}x)^2 + \frac{(-3)(-4)(-5)}{3 \times 2}(-\frac{3}{2}x)^3 + \ldots]$ | M1 |
$= \frac{1}{8} + \frac{9}{16}x + \frac{27}{16}x^2 + \frac{135}{32}x^3 + \ldots$ | A3 |
$\|x\| < \frac{2}{3}$ | B1 | (6 marks)
---\begin{enumerate}[label=(\alph*)]
\item Find the binomial expansion of $(2 - 3x)^{-3}$ in ascending powers of $x$ up to and including the term in $x^3$, simplifying each coefficient. [5]
\item State the set of values of $x$ for which your expansion is valid. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q1 [6]}}