| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Substitute expression for variable |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring standard technique: factoring out constants, applying the binomial theorem with negative index, and then making simple substitutions. Part (a) is routine C3/C4 material (5 marks), while parts (b) and (c) involve elementary substitutions (x → -x and x → x/5) that require minimal insight beyond pattern recognition. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(27(3-5x)^{-2} = 27 \times \frac{1}{9}\left(1-\frac{5}{3}x\right)^{-2}\) | B1 | Writes down \((3-5x)^{-2}\) or uses a power of \(-2\) |
| Takes out factor of \(3^{-2}\) | B1 | Implied by \(\frac{1}{9}\) or \(3\times(\ldots)\) or first term of 3 |
| \(= 3\left(1+(-2)\left(-\frac{5}{3}x\right)+\frac{(-2)(-3)}{2!}\left(-\frac{5}{3}x\right)^2+\frac{(-2)(-3)(-4)}{3!}\left(-\frac{5}{3}x\right)^3+\ldots\right)\) | M1 | Expands \((1+kx)^{-2}\), \(k\neq\pm1\) with structure for at least 2 terms correct (not including "1") |
| Two of four terms correct and simplified | A1 | Method mark must have been awarded |
| \(= 3+10x+25x^2+\frac{500}{9}x^3+\ldots\) | A1 | Fully correct simplified expansion, all on one line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(27(3+5x)^{-2} = 3-10x+25x^2-\frac{500}{9}x^3+\ldots\) | B1ft | Follow through on (a): \(A+Bx+Cx^2+Dx^3 \rightarrow A-Bx+Cx^2-Dx^3\). Must have 4 non-zero terms. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(27(3-x)^{-2} = 3+\frac{10}{5}x+\frac{25}{5^2}x^2+\frac{500}{9\times 5^3}x^3\) | M1 | Attempt to divide coefficient of \(x\) by 5, coefficient of \(x^2\) by \(5^2\), coefficient of \(x^3\) by \(5^3\), seen in at least two cases on expansion of at least 3 terms |
| \(= 3+2x+x^2+\frac{4}{9}x^3\) | A1 | Correct answer |
# Question 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $27(3-5x)^{-2} = 27 \times \frac{1}{9}\left(1-\frac{5}{3}x\right)^{-2}$ | B1 | Writes down $(3-5x)^{-2}$ or uses a power of $-2$ |
| Takes out factor of $3^{-2}$ | B1 | Implied by $\frac{1}{9}$ or $3\times(\ldots)$ or first term of 3 |
| $= 3\left(1+(-2)\left(-\frac{5}{3}x\right)+\frac{(-2)(-3)}{2!}\left(-\frac{5}{3}x\right)^2+\frac{(-2)(-3)(-4)}{3!}\left(-\frac{5}{3}x\right)^3+\ldots\right)$ | M1 | Expands $(1+kx)^{-2}$, $k\neq\pm1$ with structure for at least 2 terms correct (not including "1") |
| Two of four terms correct and simplified | A1 | **Method mark must have been awarded** |
| $= 3+10x+25x^2+\frac{500}{9}x^3+\ldots$ | A1 | Fully correct simplified expansion, all on one line |
# Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $27(3+5x)^{-2} = 3-10x+25x^2-\frac{500}{9}x^3+\ldots$ | B1ft | Follow through on (a): $A+Bx+Cx^2+Dx^3 \rightarrow A-Bx+Cx^2-Dx^3$. Must have 4 non-zero terms. |
# Question 4(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $27(3-x)^{-2} = 3+\frac{10}{5}x+\frac{25}{5^2}x^2+\frac{500}{9\times 5^3}x^3$ | M1 | Attempt to divide coefficient of $x$ by 5, coefficient of $x^2$ by $5^2$, coefficient of $x^3$ by $5^3$, seen in at least two cases on expansion of at least 3 terms |
| $= 3+2x+x^2+\frac{4}{9}x^3$ | A1 | Correct answer |
---4.
$$f ( x ) = \frac { 27 } { ( 3 - 5 x ) ^ { 2 } } \quad | x | < \frac { 3 } { 5 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient in its simplest form.\\
(5)
Use your answer to part (a) to find the series expansion in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, of
\item $g ( x ) = \frac { 27 } { ( 3 + 5 x ) ^ { 2 } } \quad | x | < \frac { 3 } { 5 }$
\item $\mathrm { h } ( x ) = \frac { 27 } { ( 3 - x ) ^ { 2 } } \quad | x | < 3$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q4 [8]}}