OCR MEI Further Pure Core 2021 November — Question 1 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeTwo linear factors in denominator
DifficultyStandard +0.3 This is a straightforward partial fractions question with two linear factors followed by a telescoping series. The partial fractions decomposition is routine, and recognizing the telescoping pattern is a standard technique taught in Further Maths. While it requires multiple steps and is from Further Maths content, it's a textbook example with no novel insight required, making it slightly easier than average overall.
Spec1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series

1
  1. Express \(\frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }\) in partial fractions.
  2. Hence find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }\), expressing the result as a single fraction.

Question 1:
AnswerMarks Guidance
1(a) 1 A B
= +
( 2 r − 1 ) ( 2 r + 1 ) 2 − r 1 2 r + 1
1 = A ( 2 r + 1 ) + B ( 2 r − 1 )
r = 1  A= 1
2 2
AnswerMarks
r = − 12  B = − 12M1
A1
A1
AnswerMarks
[3]1.1a
1.1
AnswerMarks Guidance
1.1or cover-up method
1(b) n 1 1 n ( 1 1 )
r = r −
( 2 r − 1 ) ( 2 r + 1 ) 2 2 r − 1 2 r + 1
= 1 = 1
1 (1 1 1 1 1 1 1 1 )
= − + − + ... + − + −
2 3 3 5 2 n − 3 2 n − 1 2 n − 1 2 n + 1
1( 1 )
= 1−
2 2n+1
n
=
AnswerMarks
2 n + 1M1*
M1de
p*
A1
A1
AnswerMarks
[4]1.1
1.2
1.1
AnswerMarks Guidance
1.1showing cancellation clearly SC
B3 For fully correct
summation with A = -
½ and B = ½
Question 1:
1 | (a) | 1 A B
= +
( 2 r − 1 ) ( 2 r + 1 ) 2 − r 1 2 r + 1
1 = A ( 2 r + 1 ) + B ( 2 r − 1 )
r = 1  A= 1
2 2
r = − 12  B = − 12 | M1
A1
A1
[3] | 1.1a
1.1
1.1 | or cover-up method
1 | (b) | n 1 1 n ( 1 1 )
r = r −
( 2 r − 1 ) ( 2 r + 1 ) 2 2 r − 1 2 r + 1
= 1 = 1
1 (1 1 1 1 1 1 1 1 )
= − + − + ... + − + −
2 3 3 5 2 n − 3 2 n − 1 2 n − 1 2 n + 1
1( 1 )
= 1−
2 2n+1
n
=
2 n + 1 | M1*
M1de
p*
A1
A1
[4] | 1.1
1.2
1.1
1.1 | showing cancellation clearly | SC
B3 For fully correct
summation with A = -
½ and B = ½
1
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }$ in partial fractions.
\item Hence find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) }$, expressing the result as a single fraction.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q1 [7]}}