OCR MEI Further Pure Core 2019 June — Question 16 12 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSum geometric series with complex terms
DifficultyChallenging +1.2 This is a Further Maths question requiring geometric series summation with complex exponentials and De Moivre's theorem. Part (a) is routine expansion using Euler's formula. Part (b) requires recognizing C as Re(geometric series), applying the standard formula, then algebraic manipulation using part (a). While multi-step and requiring several techniques, it follows a well-established pattern for this topic with clear signposting.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series
16
  1. Show that \(\left( 2 - \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( 2 - \mathrm { e } ^ { - \mathrm { i } \theta } \right) = 5 - 4 \cos \theta\). Series \(C\) and \(S\) are defined by \(C = \frac { 1 } { 2 } \cos \theta + \frac { 1 } { 4 } \cos 2 \theta + \frac { 1 } { 8 } \cos 3 \theta + \ldots + \frac { 1 } { 2 ^ { n } } \cos n \theta\), \(S = \frac { 1 } { 2 } \sin \theta + \frac { 1 } { 4 } \sin 2 \theta + \frac { 1 } { 8 } \sin 3 \theta + \ldots + \frac { 1 } { 2 ^ { n } } \sin n \theta\).
  2. Show that \(C = \frac { 2 ^ { n } ( 2 \cos \theta - 1 ) - 2 \cos ( n + 1 ) \theta + \cos n \theta } { 2 ^ { n } ( 5 - 4 \cos \theta ) }\).
Question 16:
AnswerMarks Guidance
16(a) (2  ei)(2  ei) = 4  2(ei+ ei) + 1
= 5  2.2cosM1 1.1b
= 5  4cos*= 5  4cos* A1
[3]
AnswerMarks Guidance
16(b) 1 1 1 1
CiS  ei e2i e3i... eni
AnswerMarks Guidance
2 4 8 2nM1 3.1a
A12.1 correct (nth term soi)
1  1 
ei 1( ei)n
2  2 
1
1 ei
AnswerMarks
2M1
A1
AnswerMarks
(4)2.1
1.1bsum of GP (condone S )
∞
correct expression
AnswerMarks Guidance
M12.1 sum of GP (condone S )
∞
AnswerMarks Guidance
A11.1b correct expression
(4)
 1 
ei 1( ei)n (2ei)
 2 
AnswerMarks Guidance
(2ei)(2ei)M1 2.1
 top and bottom by complex
conjugate
1 1
ei(2 eniei e(n1)i)
2n1 2n
AnswerMarks
54cos1 1
ei(2 eniei e(n1)i)
2n1 2n
AnswerMarks Guidance
54cos2.1 expand brackets
∞
AnswerMarks
M1expand brackets
A1
(3)
2n1ei2e(n1)i2n eni
2n(54cos)
2n1ei2e(n1)i2n eni
C Re 
2n(54cos)
AnswerMarks Guidance
 M1 2.1
∞
2n(2cos1)2cos(n1)cosn
 *
AnswerMarks
2n(54cos)2n(2cos1)2cos(n1)cosn
 *
AnswerMarks Guidance
2n(54cos)A1cao 2.2a
subsidiary denominators
(2)
[9]
M1
2.1
2.1
allow 1 slip, not on S
∞
need evidence of clearing
subsidiary denominators
AnswerMarks
Alternative solutioneni ni
e
(
M1
2
M1
A1
M1
A1
M1
A1
M1
A1
AnswerMarks Guidance
[9] )
terms of eis
sum of GP
combining fractions
expanding
expressing in cosines
NB AG
eie i e2ie 2i
1 1 1
C  ( )ei ( )...
AnswerMarks
2 2 4 2 2neni ni
e
M1
2
M1
1  1  1  1 
ei 1( ei)n ei 1( ei)n
 
2  2  2  2 
 
1 1
1 ei 1 ei
2 2
A1
M1
 1   1 
ei 1( ei)n (2ei)ei 1( ei)n (2ei)
 2   2 
AnswerMarks
(2ei)(2ei)A1
M1
A1
1 1
eiei (e(n1)ie(n1)i)2 (enieni)
2n 2n1
(2ei)(2ei)
M1
1 1
2cos cos(n1)2 cosn
2n1 2n
54cos
A1
[9]
2n(2cos1)2cos(n1)cosn
 *
2n(54cos)
subsitituting for cosines in
terms of eis
sum of GP
combining fractions
expanding
expressing in cosines
NB AG
Question 16:
16 | (a) | (2  ei)(2  ei) = 4  2(ei+ ei) + 1 | M1 | 1.1b | ei.ei = 1
= 5  2.2cos | M1 | 1.1b | ei+ ei = 2 cos used
= 5  4cos* | = 5  4cos* | A1 | 2.2a | 2.2a | NB AG | NB AG
[3]
16 | (b) | 1 1 1 1
CiS  ei e2i e3i... eni
2 4 8 2n | M1 | 3.1a | at least 2 terms
A1 | 2.1 | correct (nth term soi)
1  1 
ei 1( ei)n

2  2 

1
1 ei
2 | M1
A1
(4) | 2.1
1.1b | sum of GP (condone S )
∞
correct expression
M1 | 2.1 | sum of GP (condone S )
∞
A1 | 1.1b | correct expression
(4)
 1 
ei 1( ei)n (2ei)
 2 

(2ei)(2ei) | M1 | 2.1
 top and bottom by complex
conjugate
1 1
ei(2 eniei e(n1)i)
2n1 2n

54cos | 1 1
ei(2 eniei e(n1)i)
2n1 2n

54cos | 2.1 | expand brackets | allow 1 slip, not on S
∞
M1 | expand brackets
A1
(3)
2n1ei2e(n1)i2n eni

2n(54cos)
2n1ei2e(n1)i2n eni
C Re 
2n(54cos)
  | M1 | 2.1 | taking real part | not on S
∞
2n(2cos1)2cos(n1)cosn
 *
2n(54cos) | 2n(2cos1)2cos(n1)cosn
 *
2n(54cos) | A1cao | 2.2a | NB AG | need evidence of clearing
subsidiary denominators
(2)
[9]
M1
2.1
2.1
allow 1 slip, not on S
∞
need evidence of clearing
subsidiary denominators
Alternative solution | eni ni
e
(
M1
2
M1
A1
M1
A1
M1
A1
M1
A1
[9] |  | ) | subsitituting for cosines in
terms of eis
sum of GP
combining fractions
expanding
expressing in cosines
NB AG
eie i e2ie 2i
1 1 1
C  ( )ei ( )...
2 2 4 2 2n | eni ni
e
M1
2
M1
1  1  1  1 
ei 1( ei)n ei 1( ei)n
 
2  2  2  2 
 
1 1
1 ei 1 ei
2 2
A1
M1
 1   1 
ei 1( ei)n (2ei)ei 1( ei)n (2ei)
 2   2 

(2ei)(2ei) | A1
M1
A1
1 1
eiei (e(n1)ie(n1)i)2 (enieni)
2n 2n1

(2ei)(2ei)
M1
1 1
2cos cos(n1)2 cosn
2n1 2n

54cos
A1
[9]
2n(2cos1)2cos(n1)cosn
 *
2n(54cos)
subsitituting for cosines in
terms of eis
sum of GP
combining fractions
expanding
expressing in cosines
NB AG
16
\begin{enumerate}[label=(\alph*)]
\item Show that $\left( 2 - \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( 2 - \mathrm { e } ^ { - \mathrm { i } \theta } \right) = 5 - 4 \cos \theta$.

Series $C$ and $S$ are defined by\\
$C = \frac { 1 } { 2 } \cos \theta + \frac { 1 } { 4 } \cos 2 \theta + \frac { 1 } { 8 } \cos 3 \theta + \ldots + \frac { 1 } { 2 ^ { n } } \cos n \theta$,\\
$S = \frac { 1 } { 2 } \sin \theta + \frac { 1 } { 4 } \sin 2 \theta + \frac { 1 } { 8 } \sin 3 \theta + \ldots + \frac { 1 } { 2 ^ { n } } \sin n \theta$.
\item Show that $C = \frac { 2 ^ { n } ( 2 \cos \theta - 1 ) - 2 \cos ( n + 1 ) \theta + \cos n \theta } { 2 ^ { n } ( 5 - 4 \cos \theta ) }$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q16 [12]}}
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