| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Series expansion of rational function |
| Difficulty | Moderate -0.8 This is a straightforward application of the binomial expansion formula for negative integer powers. Students need to identify n=-3, substitute into the standard formula, and simplify coefficients—purely procedural with no problem-solving or insight required. Easier than average A-level questions. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \((2+3x)^{-3} = (2)^{-3}\left(1+\frac{3x}{2}\right)^{-3} = \frac{1}{8}\left(1+\frac{3x}{2}\right)^{-3}\) | B1 | \((2)^{-3}\) or \(\frac{1}{8}\) outside brackets, or \(\frac{1}{8}\) as constant term in expansion |
| \(= \left\{\frac{1}{8}\right\}\left[1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3+...\right]\) | M1 A1 | M1: Expands \((...+kx)^{-3}\) to give any 2 terms out of 4 terms simplified or un-simplified, with consistent \((kx)\) where \(k \neq 1\). A1: correct simplified or un-simplified expansion |
| \(= \frac{1}{8}\left[1-\frac{9}{2}x+\frac{27}{2}x^2-\frac{135}{4}x^3+...\right]\) | See notes | |
| \(= \frac{1}{8}-\frac{9}{16}x+\frac{27}{16}x^2-\frac{135}{32}x^3+...\) | A1; A1 | First A1: for \(\frac{1}{8}-\frac{9}{16}x\) (simplified fractions) or \(0.125-0.5625x\). Second A1: accept only \(\frac{27}{16}x^2-\frac{135}{32}x^3\) or \(1\frac{11}{16}x^2-4\frac{7}{32}x^3\) or \(1.6875x^2-4.21875x^3\) |
| Total | [5] | "Incorrect bracketing" \(\left\{\frac{1}{8}\right\}\left[1+(-3)\left(\frac{3x}{2}\right)+\frac{(-3)(-4)}{2!}\left(\frac{3x^2}{2}\right)+...\right]\) is M1A0 unless recovered. SC: \(\frac{1}{8}\left[1-\frac{9}{2}x;...\right]\) or \(K\left[1-\frac{9}{2}x+\frac{27}{2}x^2-\frac{135}{4}x^3+...\right]\) allow Special Case A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{8}\) or \((2)^{-3}\) | B1 | First term of expansion |
| Any two of four (un-simplified) terms correct | M1 | |
| All four (un-simplified) terms correct | A1 | |
| \(\frac{1}{8} - \frac{9}{16}x\) | A1 | |
| \(+\frac{27}{16}x^2 - \frac{135}{32}x^3\) | A1 |
## Question 1:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $(2+3x)^{-3} = (2)^{-3}\left(1+\frac{3x}{2}\right)^{-3} = \frac{1}{8}\left(1+\frac{3x}{2}\right)^{-3}$ | B1 | $(2)^{-3}$ or $\frac{1}{8}$ outside brackets, or $\frac{1}{8}$ as constant term in expansion |
| $= \left\{\frac{1}{8}\right\}\left[1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3+...\right]$ | M1 A1 | M1: Expands $(...+kx)^{-3}$ to give any 2 terms out of 4 terms simplified or un-simplified, with consistent $(kx)$ where $k \neq 1$. A1: correct simplified or un-simplified expansion |
| $= \frac{1}{8}\left[1-\frac{9}{2}x+\frac{27}{2}x^2-\frac{135}{4}x^3+...\right]$ | | See notes |
| $= \frac{1}{8}-\frac{9}{16}x+\frac{27}{16}x^2-\frac{135}{32}x^3+...$ | A1; A1 | First A1: for $\frac{1}{8}-\frac{9}{16}x$ (simplified fractions) or $0.125-0.5625x$. Second A1: accept only $\frac{27}{16}x^2-\frac{135}{32}x^3$ or $1\frac{11}{16}x^2-4\frac{7}{32}x^3$ or $1.6875x^2-4.21875x^3$ |
| **Total** | **[5]** | "Incorrect bracketing" $\left\{\frac{1}{8}\right\}\left[1+(-3)\left(\frac{3x}{2}\right)+\frac{(-3)(-4)}{2!}\left(\frac{3x^2}{2}\right)+...\right]$ is M1A0 unless recovered. SC: $\frac{1}{8}\left[1-\frac{9}{2}x;...\right]$ or $K\left[1-\frac{9}{2}x+\frac{27}{2}x^2-\frac{135}{4}x^3+...\right]$ allow Special Case A1 |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{8}$ or $(2)^{-3}$ | B1 | First term of expansion |
| Any two of four (un-simplified) terms correct | M1 | |
| All four (un-simplified) terms correct | A1 | |
| $\frac{1}{8} - \frac{9}{16}x$ | A1 | |
| $+\frac{27}{16}x^2 - \frac{135}{32}x^3$ | A1 | |
**Note:** $k = -\frac{3}{2}$ and not $\frac{3}{2}$. Terms in C need to be evaluated, so $^{-3}C_0(2)^{-3} + ^{-3}C_1(2)^{-4}(3x) + ^{-3}C_2(2)^{-5}(3x)^2 + ^{-3}C_3(2)^{-6}(3x)^3$ without further working is B0M0A0.
---\begin{enumerate}
\item Given
\end{enumerate}
$$f ( x ) = ( 2 + 3 x ) ^ { - 3 } , \quad | x | < \frac { 2 } { 3 }$$
find the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.
Give each coefficient as a simplified fraction.\\
\hfill \mbox{\textit{Edexcel C4 2013 Q1 [5]}}