Edexcel C4 2007 January — Question 1 5 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2007
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeSeries expansion of rational function
DifficultyModerate -0.3 This is a straightforward application of the binomial expansion formula for negative indices. Students need to identify n=-2, substitute into the standard formula, and simplify coefficients through basic arithmetic. It's slightly easier than average because it's a direct template application with no problem-solving required, though the negative index and fractional coefficients require careful calculation.
Spec1.04c Extend binomial expansion: rational n, |x|<1
1. $$f ( x ) = ( 2 - 5 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 5 }$$ Find the binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), as far as the term in \(x ^ { 3 }\), giving each coefficient as a simplified fraction.
(5)
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f(x) = (2-5x)^{-2} = (2)^{-2}\left(1-\frac{5x}{2}\right)^{-2} = \frac{1}{4}\left(1-\frac{5x}{2}\right)^{-2}\)B1 Takes 2 outside the bracket to give any of \((2)^{-2}\) or \(\frac{1}{4}\)
Expands \((1+x)^{-2}\) to give unsimplified \(1+(-2)(x) + \frac{(-2)(-3)}{2!}(x)^2 + \frac{(-2)(-3)(-4)}{3!}(x)^3 + \ldots\)M1 Expands \((1+x)^{-2}\) to give unsimplified \(1+(-2)(x)\)
\(= \frac{1}{4}\left\{1+(-2)\left(\frac{-5x}{2}\right) + \frac{(-2)(-3)}{2!}\left(\frac{-5x}{2}\right)^2 + \frac{(-2)(-3)(-4)}{3!}\left(\frac{-5x}{2}\right)^3 + \ldots\right\}\)A1 A correct unsimplified \(\{\ldots\}\) expansion with candidate's \((**x)\)
\(= \frac{1}{4} + \frac{5x}{4} + \frac{75x^2}{16} + \frac{125x^3}{8} + \ldots\)A1 Anything that cancels to \(\frac{1}{4} + \frac{5x}{4}\)
Simplified \(\frac{75x^2}{16} + \frac{125x^3}{8}\)A1 Simplified \(\frac{75x^2}{16} + \frac{125x^3}{8}\)
Total: 5 marks[5] 
# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (2-5x)^{-2} = (2)^{-2}\left(1-\frac{5x}{2}\right)^{-2} = \frac{1}{4}\left(1-\frac{5x}{2}\right)^{-2}$ | B1 | Takes 2 outside the bracket to give any of $(2)^{-2}$ or $\frac{1}{4}$ |
| Expands $(1+**x)^{-2}$ to give unsimplified $1+(-2)(**x) + \frac{(-2)(-3)}{2!}(**x)^2 + \frac{(-2)(-3)(-4)}{3!}(**x)^3 + \ldots$ | M1 | Expands $(1+**x)^{-2}$ to give unsimplified $1+(-2)(**x)$ |
| $= \frac{1}{4}\left\{1+(-2)\left(\frac{-5x}{2}\right) + \frac{(-2)(-3)}{2!}\left(\frac{-5x}{2}\right)^2 + \frac{(-2)(-3)(-4)}{3!}\left(\frac{-5x}{2}\right)^3 + \ldots\right\}$ | A1 | A correct unsimplified $\{\ldots\}$ expansion with candidate's $(**x)$ |
| $= \frac{1}{4} + \frac{5x}{4} + \frac{75x^2}{16} + \frac{125x^3}{8} + \ldots$ | A1 | Anything that cancels to $\frac{1}{4} + \frac{5x}{4}$ |
| Simplified $\frac{75x^2}{16} + \frac{125x^3}{8}$ | A1 | Simplified $\frac{75x^2}{16} + \frac{125x^3}{8}$ |
| **Total: 5 marks** | **[5]** | |

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1.

$$f ( x ) = ( 2 - 5 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 5 }$$

Find the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, as far as the term in $x ^ { 3 }$, giving each coefficient as a simplified fraction.\\
(5)\\

\hfill \mbox{\textit{Edexcel C4 2007 Q1 [5]}}
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