| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Multiply by polynomial |
| Difficulty | Standard +0.3 This is a standard C3/C4 binomial expansion question requiring routine application of the generalized binomial theorem with negative index, followed by straightforward algebraic manipulations (substitution and polynomial multiplication). Part (b)(i) involves simple substitution, and (b)(ii) requires multiplying the expansion by a linear polynomial—both are textbook techniques with no novel insight required. Slightly easier than average due to the mechanical nature of the steps. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{1}{(2+3x)^3} = (2+3x)^{-3} = \frac{1}{8}\left(1+\frac{3}{2}x\right)^{-3}\] | B1 | Takes out factor of \(2^{-3}\) or \(\frac{1}{8}\) or \(\frac{1}{2^3}\) (or \(0.125\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\left(1+\frac{3}{2}x\right)^{-3} = 1 + (-3)\left(\frac{3}{2}x\right) + \frac{(-3)(-4)}{2!}\left(\frac{3}{2}x\right)^2 + \ldots\] | M1 | Expands \((1+kx)^{-3}\), \(k \neq \pm 1\), with correct structure for second or third term e.g. \((-3)kx\) or \(\frac{(-3)(-4)}{2}(kx)^2\). Do not allow \(\binom{-3}{1}, \binom{-3}{2}\) for coefficients unless correct values implied by subsequent work |
| Answer | Marks | Guidance |
|---|---|---|
| \[1 + (-3)\left(\frac{3}{2}x\right) + \frac{(-3)(-4)}{2!}\left(\frac{3}{2}x\right)^2 + \ldots\] | A1 | Correct and unsimplified binomial expansion excluding the \(\left\{\frac{1}{8}\right\}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[= \frac{1}{8} - \frac{9}{16}x + \frac{27}{16}x^2\] | \(\frac{1}{8} - \frac{9}{16}x\) | A1 |
| \(\frac{27}{16}x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[(2+3x)^{-3} = 2^{-3} + (-3)\times 2^{-4}\times(3x) + \frac{(-3)(-4)}{2}\times 2^{-5}\times(3x)^2\] | B1 | First term \(2^{-3}\) |
| M1 | Correct structure for one of the other 2 terms | |
| A1 | Correct and unsimplified binomial expansion | |
| \[= \frac{1}{8} - \frac{9}{16}x + \frac{27}{16}x^2\] | \(\frac{1}{8} - \frac{9}{16}x\) | A1 |
| \(\frac{27}{16}x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[4 \times \text{"}\frac{27}{16}\text{"} = \ldots\] | M1 | \(4 \times\) their \(\frac{27}{16}\); or may expand \((2+6x)^{-3}\) fresh including their \(\frac{1}{8}\), or use their expansion from (a) with \(3x\) instead of \(\frac{3}{2}x\), and evaluates the coefficient of their \(x^2\) term |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{27}{4}\] | A1 | Allow exact equivalents e.g. \(6.75\), \(6\frac{3}{4}\). Must be identified as the required term and not just part of an expansion. |
| Answer | Marks | Guidance |
|---|---|---|
| \[4\times\text{"}\frac{27}{16}\text{"} - \left(\text{"}{-\frac{9}{16}}\text{"}\right) = \ldots\] | M1 | \(4\times\) their \(\frac{27}{16} \pm\) their \(-\frac{9}{16}\). If candidate attempts complete expansion, mark can score as long as \(x^2\) terms are collected. |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{117}{16}\] | A1 | Allow exact equivalents e.g. \(7.3125\), \(7\frac{5}{16}\). Must be identified as the required term and not just part of an expansion. |
## Question 10(a):
$$\frac{1}{(2+3x)^3} = (2+3x)^{-3} = \frac{1}{8}\left(1+\frac{3}{2}x\right)^{-3}$$ | B1 | Takes out factor of $2^{-3}$ or $\frac{1}{8}$ or $\frac{1}{2^3}$ (or $0.125$)
---
$$\left(1+\frac{3}{2}x\right)^{-3} = 1 + (-3)\left(\frac{3}{2}x\right) + \frac{(-3)(-4)}{2!}\left(\frac{3}{2}x\right)^2 + \ldots$$ | M1 | Expands $(1+kx)^{-3}$, $k \neq \pm 1$, with correct structure for second or third term e.g. $(-3)kx$ or $\frac{(-3)(-4)}{2}(kx)^2$. Do **not** allow $\binom{-3}{1}, \binom{-3}{2}$ for coefficients unless correct values implied by subsequent work
---
$$1 + (-3)\left(\frac{3}{2}x\right) + \frac{(-3)(-4)}{2!}\left(\frac{3}{2}x\right)^2 + \ldots$$ | A1 | Correct and unsimplified binomial expansion excluding the $\left\{\frac{1}{8}\right\}$
---
$$= \frac{1}{8} - \frac{9}{16}x + \frac{27}{16}x^2$$ | $\frac{1}{8} - \frac{9}{16}x$ | A1 |
$\frac{27}{16}x^2$ | A1 |
**Special Case** – if all working correct but brackets not removed:
$$= \frac{1}{8}\!\left(1 - \frac{9}{2}x + \frac{27}{2}x^2\right)$$ Score B1M1A1A1**A0**
**(5 marks total)**
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### Way 2 for 10(a):
$$(2+3x)^{-3} = 2^{-3} + (-3)\times 2^{-4}\times(3x) + \frac{(-3)(-4)}{2}\times 2^{-5}\times(3x)^2$$ | B1 | First term $2^{-3}$
| M1 | Correct structure for one of the other 2 terms
| A1 | Correct and unsimplified binomial expansion
$$= \frac{1}{8} - \frac{9}{16}x + \frac{27}{16}x^2$$ | $\frac{1}{8} - \frac{9}{16}x$ | A1 |
$\frac{27}{16}x^2$ | A1 |
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## Question 10(b)(i):
$$4 \times \text{"}\frac{27}{16}\text{"} = \ldots$$ | M1 | $4 \times$ their $\frac{27}{16}$; or may expand $(2+6x)^{-3}$ fresh including their $\frac{1}{8}$, or use their expansion from (a) with $3x$ instead of $\frac{3}{2}x$, **and evaluates the coefficient of their $x^2$ term**
---
$$\frac{27}{4}$$ | A1 | Allow exact equivalents e.g. $6.75$, $6\frac{3}{4}$. **Must be identified as the required term and not just part of an expansion.**
---
## Question 10(b)(ii):
$$4\times\text{"}\frac{27}{16}\text{"} - \left(\text{"}{-\frac{9}{16}}\text{"}\right) = \ldots$$ | M1 | $4\times$ their $\frac{27}{16} \pm$ their $-\frac{9}{16}$. If candidate attempts complete expansion, mark can score as long as $x^2$ terms are collected.
---
$$\frac{117}{16}$$ | A1 | Allow exact equivalents e.g. $7.3125$, $7\frac{5}{16}$. **Must be identified as the required term and not just part of an expansion.**
**Special Case:** If $x^2$s are included with coefficients, penalise once only at first occurrence.
**(4 marks total for 10(b))**
**(9 marks total for Q10)**\begin{enumerate}
\item (a) Use the binomial series to find the expansion of
\end{enumerate}
$$\frac { 1 } { ( 2 + 3 x ) ^ { 3 } } \quad | x | < \frac { 2 } { 3 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, giving each term as a simplified fraction.\\
(b) Hence or otherwise, find the coefficient of $x ^ { 2 }$ in the series expansion of\\
(i) $\frac { 1 } { ( 2 + 6 x ) ^ { 3 } } \quad | x | < \frac { 1 } { 3 }$\\
(ii) $\frac { 4 - x } { ( 2 + 3 x ) ^ { 3 } } \quad | x | < \frac { 2 } { 3 }$
\hfill \mbox{\textit{Edexcel C34 2019 Q10 [9]}}