OCR MEI Further Pure Core 2024 June — Question 1 4 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyModerate -0.3 This is a straightforward method of differences question where the identity is explicitly given. Students only need to verify the given expression equals 1/((r+1)(r+2)), write out the telescoping sum, and simplify. While it requires understanding of the method of differences, the question provides significant scaffolding and involves routine algebraic manipulation rather than problem-solving or insight.
Spec4.06b Method of differences: telescoping series

1 By expressing \(\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 }\) as a single fraction, find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }\) in terms of \(n\).

Question 1:
AnswerMarks
11 1 1
− =
𝑟+1 𝑟+2 (𝑟+1)(𝑟+2)
n 1 n  1 1 
so  =  − 
(r+1)(r+2) r+1 r+2
r=1 r=1
1 1 1
= − + …
2 3 3
1
…−
𝑛+2
1 1
= −
AnswerMarks
2 n + 2B1
M1*
M1dep
AnswerMarks
A11.1
2.5
2.1
AnswerMarks
2.2aDenominator may be 𝑟2+3𝑟+2. Cannot be implied.
Enough correct terms to show cancellation in their series.
First term must be correct. Fractions need not be
simplified
𝑛
isw (or )
2(𝑛+2)
AnswerMarks Guidance
Alternative methodB1 Denominator may be 𝑟2+3𝑟+2. Cannot be implied.
1 1 1
− =
𝑟+1 𝑟+2 (𝑟+1)(𝑟+2)
𝑛 𝑛+1
1 1
∑ − ∑
𝑟+1 𝑟+1
AnswerMarks Guidance
𝑟=1 𝑟=2M1* Rewriting so both series have same fraction with correct
limits
1 1
= −
AnswerMarks Guidance
1+1 (𝑛+1)+1M1dep Correct substitution of limits to leave two terms
1 1
= −
AnswerMarks Guidance
2 n + 2A1 𝑛
isw (or )
2(𝑛+2)
[4]
B1
Denominator may be 𝑟2+3𝑟+2. Cannot be implied.
Rewriting so both series have same fraction with correct
limits
Question 1:
1 | 1 1 1
− =
𝑟+1 𝑟+2 (𝑟+1)(𝑟+2)
n 1 n  1 1 
so  =  − 
(r+1)(r+2) r+1 r+2
r=1 r=1
1 1 1
= − + …
2 3 3
1
…−
𝑛+2
1 1
= −
2 n + 2 | B1
M1*
M1dep
A1 | 1.1
2.5
2.1
2.2a | Denominator may be 𝑟2+3𝑟+2. Cannot be implied.
Enough correct terms to show cancellation in their series.
First term must be correct. Fractions need not be
simplified
𝑛
isw (or )
2(𝑛+2)
Alternative method | B1 | Denominator may be 𝑟2+3𝑟+2. Cannot be implied.
1 1 1
− =
𝑟+1 𝑟+2 (𝑟+1)(𝑟+2)
𝑛 𝑛+1
1 1
∑ − ∑
𝑟+1 𝑟+1
𝑟=1 𝑟=2 | M1* | Rewriting so both series have same fraction with correct
limits
1 1
= −
1+1 (𝑛+1)+1 | M1dep | Correct substitution of limits to leave two terms
1 1
= −
2 n + 2 | A1 | 𝑛
isw (or )
2(𝑛+2)
[4]
B1
Denominator may be 𝑟2+3𝑟+2. Cannot be implied.
Rewriting so both series have same fraction with correct
limits
1 By expressing $\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 }$ as a single fraction, find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }$ in terms of $n$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q1 [4]}}