Edexcel C4 — Question 3 9 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions then binomial expansion
DifficultyStandard +0.3 This is a standard C4 question combining two routine techniques: partial fractions decomposition of a quadratic denominator, followed by binomial expansion of each fraction. Both steps are textbook procedures with no novel insight required, though the multi-step nature and algebraic manipulation place it slightly above average difficulty.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1
3. (a) Express \(\frac { 2 + 20 x } { 1 + 2 x - 8 x ^ { 2 } }\) as a sum of partial fractions.
(b) Hence find the series expansion of \(\frac { 2 + 20 x } { 1 + 2 x - 8 x ^ { 2 } } , | x | < \frac { 1 } { 4 }\), in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\), simplifying each coefficient.
3. continued
AnswerMarks Guidance
(a) \(\frac{2 + 20x}{1 + 2x - 8x^2} = \frac{2 + 20x}{(1-2x)(1+4x)} = \frac{A}{1-2x} + \frac{B}{1+4x}\)B1
\(2 + 20x \equiv A(1 + 4x) + B(1 - 2x)\)M1
\(x = \frac{1}{2} \Rightarrow 12 = 3A \Rightarrow A = 4\)A1
\(x = -\frac{1}{4} \Rightarrow -3 = \frac{3}{2}B \Rightarrow B = -2\)A1
\(\frac{2 + 20x}{1 + 2x - 8x^2} = \frac{4}{1-2x} - \frac{2}{1+4x}\)A1
(b) \(\frac{2 + 20x}{1+2x-8x^2} = 4(1-2x)^{-1} - 2(1+4x)^{-1}\)M1
\((1 - 2x)^{-1} = 1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + \frac{(-1)(-2)(-3)}{3 \cdot 2}(-2x)^3 + \ldots = 1 + 2x + 4x^2 + 8x^3 + \ldots\)M1 A1
\((1 + 4x)^{-1} = 1 + (-1)(4x) + \frac{(-1)(-2)}{2}(4x)^2 + \frac{(-1)(-2)(-3)}{3 \cdot 2}(4x)^3 + \ldots = 1 - 4x + 16x^2 - 64x^3 + \ldots\)A1
\(\frac{2 + 20x}{1+2x-8x^2} = 4(1 + 2x + 4x^2 + 8x^3 + \ldots) - 2(1 - 4x + 16x^2 - 64x^3 + \ldots)\)M1
\(= 2 + 16x - 16x^2 + 160x^3 + \ldots\)A1 (9) 
**(a)** $\frac{2 + 20x}{1 + 2x - 8x^2} = \frac{2 + 20x}{(1-2x)(1+4x)} = \frac{A}{1-2x} + \frac{B}{1+4x}$ | B1 |

$2 + 20x \equiv A(1 + 4x) + B(1 - 2x)$ | M1 |

$x = \frac{1}{2} \Rightarrow 12 = 3A \Rightarrow A = 4$ | A1 |

$x = -\frac{1}{4} \Rightarrow -3 = \frac{3}{2}B \Rightarrow B = -2$ | A1 |

$\frac{2 + 20x}{1 + 2x - 8x^2} = \frac{4}{1-2x} - \frac{2}{1+4x}$ | A1 |

**(b)** $\frac{2 + 20x}{1+2x-8x^2} = 4(1-2x)^{-1} - 2(1+4x)^{-1}$ | M1 |

$(1 - 2x)^{-1} = 1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + \frac{(-1)(-2)(-3)}{3 \cdot 2}(-2x)^3 + \ldots = 1 + 2x + 4x^2 + 8x^3 + \ldots$ | M1 A1 |

$(1 + 4x)^{-1} = 1 + (-1)(4x) + \frac{(-1)(-2)}{2}(4x)^2 + \frac{(-1)(-2)(-3)}{3 \cdot 2}(4x)^3 + \ldots = 1 - 4x + 16x^2 - 64x^3 + \ldots$ | A1 |

$\frac{2 + 20x}{1+2x-8x^2} = 4(1 + 2x + 4x^2 + 8x^3 + \ldots) - 2(1 - 4x + 16x^2 - 64x^3 + \ldots)$ | M1 |

$= 2 + 16x - 16x^2 + 160x^3 + \ldots$ | A1 | (9)
3. (a) Express $\frac { 2 + 20 x } { 1 + 2 x - 8 x ^ { 2 } }$ as a sum of partial fractions.\\
(b) Hence find the series expansion of $\frac { 2 + 20 x } { 1 + 2 x - 8 x ^ { 2 } } , | x | < \frac { 1 } { 4 }$, in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, simplifying each coefficient.\\
3. continued\\

\hfill \mbox{\textit{Edexcel C4  Q3 [9]}}
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