AQA Further Paper 2 Specimen — Question 15 10 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeDeduce related series from given series
DifficultyChallenging +1.3 This is a Further Maths question involving complex numbers and series, requiring multiple techniques (expanding complex conjugates, geometric series with complex terms, separating real/imaginary parts). While it requires several steps and coordination across parts, each individual step follows standard Further Maths procedures without requiring novel insight—the path is clearly signposted through the structured parts.
Spec1.04j Sum to infinity: convergent geometric series |r|<14.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)
  1. Show that \((1-\frac{1}{4}e^{2i\theta})(1-\frac{1}{4}e^{-2i\theta}) = \frac{1}{16}(17-8\cos 2\theta)\) [3 marks]
  2. Given that the series \(e^{2i\theta} + \frac{1}{4}e^{4i\theta} + \frac{1}{16}e^{6i\theta} + \frac{1}{64}e^{8i\theta} + \ldots\) has a sum to infinity, express this sum to infinity in terms of \(e^{2i\theta}\) [2 marks]
  3. Hence show that \(\sum_{n=1}^{\infty} \frac{1}{4^{n-1}} \cos 2n\theta = \frac{16\cos 2\theta - 4}{17 - 8\cos 2\theta}\) [4 marks]
  4. Deduce a similar expression for \(\sum_{n=1}^{\infty} \frac{1}{4^{n-1}} \sin 2n\theta\) [1 mark]
Question 15:
AnswerMarks
15(a)Commences an argument by
correctly expanding brackets and
AnswerMarks Guidance
simplifying final term to 1/16AO1.1a M1
(1 e2i)(1 e2i)
4 4
1 1 1
1 e2i e2i
4 4 16
17 1 1
  (cos 2isin 2) (cos 2isin 2)
16 4 4
17 1
  cos 2
16 2
1
 (178cos 2)
16
Substitutes correctly for both
e2iand e2i in terms of cos 2
and sin 2 (seen anywhere in
AnswerMarks Guidance
solution)AO1.1b B1
Completes argument and reaches
stated result by collecting terms
and simplifying correctly, no
AnswerMarks Guidance
errors in working seen AGAO2.1 R1
(b)Identifies series as a geometric
series and states first term and
AnswerMarks Guidance
common ratio correctlyAO1.1b B1
1
and common ratio a  e2i
4
a e2i
S  
 1r 1
1 e2i
4
States and uses sum to infinity
formula correctly
FT incorrect values for first term
AnswerMarks Guidance
and common ratioAO1.1b B1F
QMarking Instructions AO
(c)Deduces that the series in part (c)
is related to the real part of the
AnswerMarks Guidance
series in part (b)AO2.2a R1
1 1 1
e2i e4i e6i e8i....
4 16 64
Using result from previous part
1
(1 e2i)
e2i e2i
4
 
1 1 1
1 e2i (1 e2i) (1 e2i)
4 4 4
1
e2i
4
1 1
(1 e2i)(1 e2i)
4 4
1
cos 2 isin 2
4
1
(178cos 2)
16
Real part =
1
cos 2
4 16cos 24
1 178cos 2
(178cos 2)
16
Selects an appropriate method by
using the result in part (b) and
multiplying appropriately to
AnswerMarks Guidance
realise the denominatorAO3.1a M1
Substitutes to obtain an
expression with cosines and
sines only – using part (a)
FT incorrect sum to infinity
AnswerMarks Guidance
provided M1 has been awardedAO1.1b A1F
Identifies the real part and
correctly completes the argument
to reach the stated result.
Only award for an error-free fully
AnswerMarks Guidance
correct solutionAO2.1 R1
(d)Identifies the imaginary part and
states the correct expressionAO2.2a R1
given series hence
sin 2 16sin 2
1 178cos 2
(178cos 2)
16
AnswerMarks Guidance
Total10
QMarking Instructions AO 
Question 15:
--- 15(a) ---
15(a) | Commences an argument by
correctly expanding brackets and
simplifying final term to 1/16 | AO1.1a | M1 | 1 1
(1 e2i)(1 e2i)
4 4
1 1 1
1 e2i e2i
4 4 16
17 1 1
  (cos 2isin 2) (cos 2isin 2)
16 4 4
17 1
  cos 2
16 2
1
 (178cos 2)
16
Substitutes correctly for both
e2iand e2i in terms of cos 2
and sin 2 (seen anywhere in
solution) | AO1.1b | B1
Completes argument and reaches
stated result by collecting terms
and simplifying correctly, no
errors in working seen AG | AO2.1 | R1
(b) | Identifies series as a geometric
series and states first term and
common ratio correctly | AO1.1b | B1 | Geometric series with first term r e2i
1
and common ratio a  e2i
4
a e2i
S  
 1r 1
1 e2i
4
States and uses sum to infinity
formula correctly
FT incorrect values for first term
and common ratio | AO1.1b | B1F
Q | Marking Instructions | AO | Marks | Typical Solution
(c) | Deduces that the series in part (c)
is related to the real part of the
series in part (b) | AO2.2a | R1 | Series stated = real part of the series
1 1 1
e2i e4i e6i e8i....
4 16 64
Using result from previous part
1
(1 e2i)
e2i e2i
4
 
1 1 1
1 e2i (1 e2i) (1 e2i)
4 4 4
1
e2i
4

1 1
(1 e2i)(1 e2i)
4 4
1
cos 2 isin 2
4
1
(178cos 2)
16
Real part =
1
cos 2
4 16cos 24

1 178cos 2
(178cos 2)
16
Selects an appropriate method by
using the result in part (b) and
multiplying appropriately to
realise the denominator | AO3.1a | M1
Substitutes to obtain an
expression with cosines and
sines only – using part (a)
FT incorrect sum to infinity
provided M1 has been awarded | AO1.1b | A1F
Identifies the real part and
correctly completes the argument
to reach the stated result.
Only award for an error-free fully
correct solution | AO2.1 | R1
(d) | Identifies the imaginary part and
states the correct expression | AO2.2a | R1 | Required series = imaginary part of the
given series hence
sin 2 16sin 2

1 178cos 2
(178cos 2)
16
Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Show that $(1-\frac{1}{4}e^{2i\theta})(1-\frac{1}{4}e^{-2i\theta}) = \frac{1}{16}(17-8\cos 2\theta)$
[3 marks]

\item Given that the series $e^{2i\theta} + \frac{1}{4}e^{4i\theta} + \frac{1}{16}e^{6i\theta} + \frac{1}{64}e^{8i\theta} + \ldots$ has a sum to infinity, express this sum to infinity in terms of $e^{2i\theta}$
[2 marks]

\item Hence show that $\sum_{n=1}^{\infty} \frac{1}{4^{n-1}} \cos 2n\theta = \frac{16\cos 2\theta - 4}{17 - 8\cos 2\theta}$
[4 marks]

\item Deduce a similar expression for $\sum_{n=1}^{\infty} \frac{1}{4^{n-1}} \sin 2n\theta$
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2  Q15 [10]}}
This paper (17 questions)
View full paper
Q1 1 Q2 3 Q3 3 Q4 4 Q5 4 Q6 5 Q7 5 Q7 2 Q8 5 Q9 6 Q10 8 Q11 8 Q12 11 Q13 7 Q14 9 Q15 10 Q16 9