OCR Further Pure Core 2 2023 June — Question 7 8 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypeComplex partial fractions with multiple techniques
DifficultyChallenging +1.8 This is a Further Maths question requiring partial fractions decomposition, telescoping series manipulation, and limit evaluation. While it involves multiple techniques (partial fractions, series summation, limits), the path is relatively standard for FM students: decompose the fraction, identify telescoping terms, and apply the given Basel problem result. The algebraic manipulation is moderately involved but follows established patterns for this topic.
Spec4.06b Method of differences: telescoping series
7 In this question you must show detailed reasoning.
  1. Show that $$\sum _ { r = 1 } ^ { n } \frac { 5 r + 6 } { r ^ { 3 } + r ^ { 2 } } = \frac { a } { n + 1 } + b + c \sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are integers whose values are to be determined. You are given that \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } }\) exists and is equal to \(\frac { 1 } { 6 } \pi ^ { 2 }\).
  2. Show that \(\sum _ { r = 1 } ^ { \infty } \frac { 5 r + 6 } { r ^ { 3 } + r ^ { 2 } }\) exists and is equal to \(( \pi - 1 ) ( \pi + 1 )\).
Question 7:
AnswerMarks Guidance
7(a) DR
5r+6 A B C
= + +
AnswerMarks Guidance
r3+r2 r r2 r+1*M1 3.1a
solution.
Allow extraneous terms only if
AnswerMarks
stated/evaluated as 0A r + B C
Accept + but not with
r 2 r + 1
D
additional unless recovered later
r
A r ( r + 1 ) + B ( r + 1 ) + C r 2
=
r 2 ( r + 1 )
AnswerMarks Guidance
C o n s i d e r r = − 1*M1 1.1
method to find coefficients
(eg appropriate choice of r or
comparing coefficients in
AnswerMarks
r2, r or r0).Indep of 1st M1 if from PF terms
involving 3+ unknowns and their
factors seen in a denominator
AnswerMarks Guidance
C = 1A1 1.1
coefficient
AnswerMarks Guidance
r = 0 =>B = 6 and A + C = 0 =>A = –1A1 1.1
method
n 5 r3 + 6 n  − 1 62 1 
 r =  + +
r + 2 r r r r + 1 
= 1 = r 1
n 62 n  1 1 
= r +   −
r r + 1 r
= 1 = r 1
n 62 1 1 1 1 1 1
= r + − + − + − + ...
r 2 1 3 2 4 3
= 1
1 1 1 1
... + − + −
AnswerMarks Guidance
n n − 1 n + 1 ndep*M1 1.1
in form in which cancellation
AnswerMarks
pattern is clearr = ' r + 1
n 1 n 1 nr + 1 1 n 1
  − r = − 
r + 1 r r ' r
= r 1 = 1 =' 2 1
1 n  1 1  1 1
= + r − − = − 1
n + 1 r r 1 n + 1
= 2
1 n 1
= − 1 + 6 r
n + 1 r 2
= 1
AnswerMarks Guidance
soa = 1, b = –1, c = 6A1 2.2a
detail.
AnswerMarks
Allow embedded answersMinimum of first and last
cancellation terms shown
(If algebraic approach, must see
full argument)
[6]
AnswerMarks
(b)1
lim =0
AnswerMarks Guidance
n→n+1M1 3.1a
1
tends to infinity of term
AnswerMarks
𝑛+11
not
infinity
5 r3 62 n 5 r3 62  +  + 
li m  = 
r r n r r + →  +
r 1 r 1 = =
1 n 1  
li m 1 6 2 1  = − +  = −
n n 1 2 r →  +
r 1 =
AnswerMarks Guidance
( 1 ) ( 1 ) .   = − +A1 2.2a
intermediate working must be
AnswerMarks
shownCondone poor, but clear limit
notation,
[2]
Question 7:
7 | (a) | DR
5r+6 A B C
= + +
r3+r2 r r2 r+1 | *M1 | 3.1a | Correct PF expansion used in
solution.
Allow extraneous terms only if
stated/evaluated as 0 | A r + B C
Accept + but not with
r 2 r + 1
D
additional unless recovered later
r
A r ( r + 1 ) + B ( r + 1 ) + C r 2
=
r 2 ( r + 1 )
C o n s i d e r r = − 1 | *M1 | 1.1 | Recombining and use valid
method to find coefficients
(eg appropriate choice of r or
comparing coefficients in
r2, r or r0). | Indep of 1st M1 if from PF terms
involving 3+ unknowns and their
factors seen in a denominator
C = 1 | A1 | 1.1 | Any one correct non-zero
coefficient
r = 0 =>B = 6 and A + C = 0 =>A = –1 | A1 | 1.1 | Other two coefficients by valid
method
n 5 r3 + 6 n  − 1 62 1 
 r =  + +
r + 2 r r r r + 1 
= 1 = r 1
n 62 n  1 1 
= r +   −
r r + 1 r
= 1 = r 1
n 62 1 1 1 1 1 1
= r + − + − + − + ...
r 2 1 3 2 4 3
= 1
1 1 1 1
... + − + −
n n − 1 n + 1 n | dep*M1 | 1.1 | Separating and expressing sum
in form in which cancellation
pattern is clear | r = ' r + 1
n 1 n 1 nr + 1 1 n 1
  − r = − 
r + 1 r r ' r
= r 1 = 1 =' 2 1
1 n  1 1  1 1
= + r − − = − 1
n + 1 r r 1 n + 1
= 2
1 n 1
= − 1 + 6 r
n + 1 r 2
= 1
soa = 1, b = –1, c = 6 | A1 | 2.2a | Complete argument with all
detail.
Allow embedded answers | Minimum of first and last
cancellation terms shown
(If algebraic approach, must see
full argument)
[6]
(b) | 1
lim =0
n→n+1 | M1 | 3.1a | AG. Considering the limit as n
1
tends to infinity of term
𝑛+1 | 1
not
infinity
5 r3 62 n 5 r3 62  +  + 
li m  = 
r r n r r + →  +
r 1 r 1 = =
1 n 1  
li m 1 6 2 1  = − +  = −
n n 1 2 r →  +
r 1 =
( 1 ) ( 1 ) .   = − + | A1 | 2.2a | AG. Joined up argument Some
intermediate working must be
shown | Condone poor, but clear limit
notation,
[2]
7 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Show that

$$\sum _ { r = 1 } ^ { n } \frac { 5 r + 6 } { r ^ { 3 } + r ^ { 2 } } = \frac { a } { n + 1 } + b + c \sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } }$$

where $a$, $b$ and $c$ are integers whose values are to be determined.

You are given that $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } }$ exists and is equal to $\frac { 1 } { 6 } \pi ^ { 2 }$.
\item Show that $\sum _ { r = 1 } ^ { \infty } \frac { 5 r + 6 } { r ^ { 3 } + r ^ { 2 } }$ exists and is equal to $( \pi - 1 ) ( \pi + 1 )$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2023 Q7 [8]}}
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