| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown constant from coefficient |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial series requiring students to expand (2-3x)^{-3}, then multiply by a linear numerator and match coefficients. While it involves multiple steps, each step follows a standard procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2^{-3}\) or \(\frac{1}{2^3}\) or \(0.125\) | B1 | For taking out factor of \(2^{-3}\) |
| \(\frac{1}{(2-3x)^3} = \frac{1}{2^3}\left(1-\frac{3x}{2}\right)^{-3} = \frac{1}{8}\left(1+(-3)\times\left(-\frac{3x}{2}\right)+\frac{-3\times-4}{2!}\times\left(-\frac{3x}{2}\right)^2+...\right)\) | M1A1 | M1: binomial expansion with index \(-3\), structure of at least one other term correct. A1: correct unsimplified form |
| \(= \frac{1}{8} + \frac{9}{16}x + \frac{27}{16}x^2 + ...\) | A1; A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4+kx}{(2-3x)^3} = (4+kx)\left(\frac{1}{8}+\frac{9}{16}x+\frac{27}{16}x^2+...\right)\) | ||
| Compares \(x^2\) terms: \(\frac{27}{4}+\frac{9k}{16} = \frac{81}{16} \Rightarrow k=...\) | M1 | |
| \(k=-3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Compares \(x\) terms: \(\frac{9}{4}+\frac{1}{8}\times(-3) = A \Rightarrow A=...\) | M1 | |
| \(A = \frac{15}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{8} + \frac{9}{16}x\) | A1 | First two terms correct and simplified |
| \(+\frac{27}{16}x^2\) | A1 | Third term correct and simplified. Allow from expansion using \(\frac{3x}{2}\) rather than \(-\frac{3x}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4 \times \frac{27}{16} + \frac{9}{16}k = \frac{81}{16} \Rightarrow k = \ldots\) | M1 | Finds sum of coefficients of two \(x^2\) terms, sets equal to \(\frac{81}{16}\), proceeds to find \(k\) |
| \(k = -3\) | A1 | Must come from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 4 \times \frac{9}{16} + \frac{1}{8}k\) | M1 | Finds sum of coefficients of two \(x\) terms using value of \(k\), proceeds to find \(A\) |
| \(A = \frac{15}{8}\) | A1 | oe e.g. 1.875. If \(k=-3\) obtained fortuitously in (b), allow \(A=\frac{15}{8}\) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2^{-3}$ or $\frac{1}{2^3}$ or $0.125$ | B1 | For taking out factor of $2^{-3}$ |
| $\frac{1}{(2-3x)^3} = \frac{1}{2^3}\left(1-\frac{3x}{2}\right)^{-3} = \frac{1}{8}\left(1+(-3)\times\left(-\frac{3x}{2}\right)+\frac{-3\times-4}{2!}\times\left(-\frac{3x}{2}\right)^2+...\right)$ | M1A1 | M1: binomial expansion with index $-3$, structure of at least one other term correct. A1: correct unsimplified form |
| $= \frac{1}{8} + \frac{9}{16}x + \frac{27}{16}x^2 + ...$ | A1; A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4+kx}{(2-3x)^3} = (4+kx)\left(\frac{1}{8}+\frac{9}{16}x+\frac{27}{16}x^2+...\right)$ | | |
| Compares $x^2$ terms: $\frac{27}{4}+\frac{9k}{16} = \frac{81}{16} \Rightarrow k=...$ | M1 | |
| $k=-3$ | A1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Compares $x$ terms: $\frac{9}{4}+\frac{1}{8}\times(-3) = A \Rightarrow A=...$ | M1 | |
| $A = \frac{15}{8}$ | A1 | |
# Question 7 (Binomial Expansion):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{8} + \frac{9}{16}x$ | A1 | First two terms correct and simplified |
| $+\frac{27}{16}x^2$ | A1 | Third term correct and simplified. Allow from expansion using $\frac{3x}{2}$ rather than $-\frac{3x}{2}$ |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 \times \frac{27}{16} + \frac{9}{16}k = \frac{81}{16} \Rightarrow k = \ldots$ | M1 | Finds sum of coefficients of two $x^2$ terms, sets equal to $\frac{81}{16}$, proceeds to find $k$ |
| $k = -3$ | A1 | Must come from correct work |
**(c)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 4 \times \frac{9}{16} + \frac{1}{8}k$ | M1 | Finds sum of coefficients of two $x$ terms using value of $k$, proceeds to find $A$ |
| $A = \frac{15}{8}$ | A1 | oe e.g. 1.875. If $k=-3$ obtained fortuitously in (b), allow $A=\frac{15}{8}$ |
---\begin{enumerate}
\item (a) Use the binomial series to expand
\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.
$$f ( x ) = \frac { 4 + k x } { ( 2 - 3 x ) ^ { 3 } } \quad \text { where } k \text { is a constant and } | x | < \frac { 2 } { 3 }$$
Given that the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, is
$$\frac { 1 } { 2 } + A x + \frac { 81 } { 16 } x ^ { 2 } + \cdots$$
where $A$ is a constant,\\
(b) find the value of $k$,\\
(c) find the value of $A$.\\
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\hfill \mbox{\textit{Edexcel C34 2017 Q7 [9]}}