OCR MEI Further Pure Core Specimen — Question 5 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeTwo linear factors in denominator
DifficultyStandard +0.8 This is a Further Maths question requiring partial fractions followed by telescoping series summation. Part (i) is routine, but part (ii) requires recognizing the telescoping pattern and carefully tracking which terms survive—a technique beyond standard A-level that requires methodical algebraic manipulation and insight into series behavior.
Spec4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series
  1. Express \(\frac{2}{(r+1)(r+3)}\) in partial fractions. [2]
  2. Hence find \(\sum_{r=1}^{n} \frac{1}{(r+1)(r+3)}\), expressing your answer as a single fraction. [5]
Question 5:
AnswerMarks Guidance
5(i) 2 A B
(cid:32) (cid:14)
(cid:11)r(cid:14)1(cid:12)(cid:11)r(cid:14)3(cid:12) (cid:11)r(cid:14)1(cid:12) (cid:11)r(cid:14)3(cid:12)
2(cid:32)A(r(cid:14)3)(cid:14)B(r(cid:14)1)
A = 1, B = –1
2 1 1
(cid:32) (cid:16)
AnswerMarks
(cid:11)r(cid:14)1(cid:12)(cid:11)r(cid:14)3(cid:12) (cid:11)r(cid:14)1(cid:12) (cid:11)r(cid:14)3(cid:12)M1
A1
AnswerMarks
[2]1.1
1.1Correct form plus some
progress
n
AnswerMarks Guidance
5(ii) (cid:173)(cid:167)1 1(cid:183) (cid:167)1 1(cid:183) (cid:189)
(cid:16) (cid:14) (cid:16) (cid:14).....
(cid:176)(cid:168) (cid:184) (cid:168) (cid:184) (cid:176)
n 1 1(cid:176)(cid:169)2 4(cid:185) (cid:169)3 5(cid:185) (cid:176)
(cid:166) (cid:32) (cid:174) (cid:190)
r(cid:32)1 (r(cid:14)1)(r(cid:14)3) 2 (cid:176) (cid:14) (cid:167)1 (cid:16) 1 (cid:183) (cid:14) (cid:167) 1 (cid:16) 1 (cid:183)(cid:176)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:176) (cid:175) (cid:169)n n(cid:14)2(cid:185) (cid:169)n(cid:14)1 n(cid:14)3(cid:185) (cid:176) (cid:191)
1(cid:167)1 1 1 1 (cid:183)
(cid:32) (cid:168) (cid:14) (cid:16) (cid:16) (cid:184)
2(cid:169)2 3 n(cid:14)2 n(cid:14)3(cid:185)
3(n(cid:14)2)(n(cid:14)3)(cid:14)2(n(cid:14)2)(n(cid:14)3)(cid:16)6(n(cid:14)3)(cid:16)6(n(cid:14)2)
12(n(cid:14)2)(n(cid:14)3)
5n2 (cid:14)25n(cid:14)30(cid:16)12n(cid:16)30
(cid:32)
12(n(cid:14)2)(n(cid:14)3) e
5n2 (cid:14)13n
(cid:32) p
12(cid:11)n(cid:14)2(cid:12)(cid:11)n(cid:14)3(cid:12)
AnswerMarks
SM1
A1
A1
M1
c
A1
AnswerMarks
[5]3.1a
2.1
2.1
m
2.1
i
AnswerMarks
1.1Write out series to identify:
Cancelling terms,
denominators 4, 5, … n+1
Ignore absence of initial 1 .
2
Nnon-cancelling terms at the
beginning.
eIgnore absence of initial 1 .
2
Non-cancelling terms at the
end:
Ignore absence of initial 1 .
2
Attempt at writing as single
fraction, with product of their
linear terms.
Completely correct
argument; each of numerator
and denominator can be
either factorised or
multiplied out.
Question 5:
5 | (i) | 2 A B
(cid:32) (cid:14)
(cid:11)r(cid:14)1(cid:12)(cid:11)r(cid:14)3(cid:12) (cid:11)r(cid:14)1(cid:12) (cid:11)r(cid:14)3(cid:12)
2(cid:32)A(r(cid:14)3)(cid:14)B(r(cid:14)1)
A = 1, B = –1
2 1 1
(cid:32) (cid:16)
(cid:11)r(cid:14)1(cid:12)(cid:11)r(cid:14)3(cid:12) (cid:11)r(cid:14)1(cid:12) (cid:11)r(cid:14)3(cid:12) | M1
A1
[2] | 1.1
1.1 | Correct form plus some
progress
n
5 | (ii) | (cid:173)(cid:167)1 1(cid:183) (cid:167)1 1(cid:183) (cid:189)
(cid:16) (cid:14) (cid:16) (cid:14).....
(cid:176)(cid:168) (cid:184) (cid:168) (cid:184) (cid:176)
n 1 1(cid:176)(cid:169)2 4(cid:185) (cid:169)3 5(cid:185) (cid:176)
(cid:166) (cid:32) (cid:174) (cid:190)
r(cid:32)1 (r(cid:14)1)(r(cid:14)3) 2 (cid:176) (cid:14) (cid:167)1 (cid:16) 1 (cid:183) (cid:14) (cid:167) 1 (cid:16) 1 (cid:183)(cid:176)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:176) (cid:175) (cid:169)n n(cid:14)2(cid:185) (cid:169)n(cid:14)1 n(cid:14)3(cid:185) (cid:176) (cid:191)
1(cid:167)1 1 1 1 (cid:183)
(cid:32) (cid:168) (cid:14) (cid:16) (cid:16) (cid:184)
2(cid:169)2 3 n(cid:14)2 n(cid:14)3(cid:185)
3(n(cid:14)2)(n(cid:14)3)(cid:14)2(n(cid:14)2)(n(cid:14)3)(cid:16)6(n(cid:14)3)(cid:16)6(n(cid:14)2)
12(n(cid:14)2)(n(cid:14)3)
5n2 (cid:14)25n(cid:14)30(cid:16)12n(cid:16)30
(cid:32)
12(n(cid:14)2)(n(cid:14)3) e
5n2 (cid:14)13n
(cid:32) p
12(cid:11)n(cid:14)2(cid:12)(cid:11)n(cid:14)3(cid:12)
S | M1
A1
A1
M1
c
A1
[5] | 3.1a
2.1
2.1
m
2.1
i
1.1 | Write out series to identify:
Cancelling terms,
denominators 4, 5, … n+1
Ignore absence of initial 1 .
2
Nnon-cancelling terms at the
beginning.
eIgnore absence of initial 1 .
2
Non-cancelling terms at the
end:
Ignore absence of initial 1 .
2
Attempt at writing as single
fraction, with product of their
linear terms.
Completely correct
argument; each of numerator
and denominator can be
either factorised or
multiplied out.
\begin{enumerate}[label=(\roman*)]
\item Express $\frac{2}{(r+1)(r+3)}$ in partial fractions. [2]

\item Hence find $\sum_{r=1}^{n} \frac{1}{(r+1)(r+3)}$, expressing your answer as a single fraction. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q5 [7]}}
This paper (16 questions)
View full paper
Q1 3 Q2 6 Q3 6 Q4 5 Q5 7 Q6 6 Q7 11 Q8 5 Q9 7 Q10 9 Q11 9 Q12 13 Q13 13 Q14 18 Q15 8 Q16 18