OCR Further Pure Core 2 Specimen — Question 3 4 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyStandard +0.3 This is a standard telescoping series question requiring recognition of the pattern and simplification. Part (i) involves writing out terms to see cancellation (a routine Further Maths technique), and part (ii) is immediate once part (i) is complete. While it's Further Maths content, the execution is mechanical with no novel insight required, making it slightly easier than average overall.
Spec4.06b Method of differences: telescoping series

\begin{enumerate}[label=(\roman*)] \item Find \(\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+2}\right)\). [3] \item What does the sum in part (i) tend to as \(n \to \infty\)? Justify your answer. [1]
Question 3:
AnswerMarks Guidance
3(i) (cid:167) 1(cid:183) (cid:167)1 1(cid:183) (cid:167)1 1(cid:183)
1(cid:16) (cid:14) (cid:16) (cid:14) (cid:16) (cid:14)...
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) 3(cid:185) (cid:169)2 4(cid:185) (cid:169)3 5(cid:185)
(cid:167) 1 1 (cid:183) (cid:167)1 1 (cid:183)
(cid:14) (cid:16) (cid:14) (cid:16)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)n(cid:16)1 n(cid:14)1(cid:185) (cid:169)n n(cid:14)2(cid:185)
Cancelling of inner terms
3 1 1
(cid:16) (cid:16)
AnswerMarks
2 n(cid:14)1 n(cid:14)2M1
M1
A1
AnswerMarks
[3]2.1
1.1
AnswerMarks
1.1At least first and last terms in full
n
AnswerMarks Guidance
3(ii) 1 1 3
(cid:111)0 and (cid:111)0, hence sum (cid:111)
AnswerMarks Guidance
n(cid:14)1 n(cid:14)2 2E1
[1]2.4
me Justification required
E0 for unsupported answer
Question 3:
3 | (i) | (cid:167) 1(cid:183) (cid:167)1 1(cid:183) (cid:167)1 1(cid:183)
1(cid:16) (cid:14) (cid:16) (cid:14) (cid:16) (cid:14)...
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) 3(cid:185) (cid:169)2 4(cid:185) (cid:169)3 5(cid:185)
(cid:167) 1 1 (cid:183) (cid:167)1 1 (cid:183)
(cid:14) (cid:16) (cid:14) (cid:16)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)n(cid:16)1 n(cid:14)1(cid:185) (cid:169)n n(cid:14)2(cid:185)
Cancelling of inner terms
3 1 1
(cid:16) (cid:16)
2 n(cid:14)1 n(cid:14)2 | M1
M1
A1
[3] | 2.1
1.1
1.1 | At least first and last terms in full
n
3 | (ii) | 1 1 3
(cid:111)0 and (cid:111)0, hence sum (cid:111)
n(cid:14)1 n(cid:14)2 2 | E1
[1] | 2.4
m | e | Justification required
E0 for unsupported answer
\begin{enumerate}[label=(\roman*)]
\item Find $\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+2}\right)$. [3]

\item What does the sum in part (i) tend to as $n \to \infty$? Justify your answer. [1]
</end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2  Q3 [4]}}