OCR Further Pure Core 2 2019 June — Question 1 7 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeInfinite series convergence and sum
DifficultyStandard +0.3 This is a standard telescoping series question requiring partial fractions decomposition and recognizing cancellation patterns. While it involves multiple steps (factorizing, partial fractions, summing, taking limit), these are all routine techniques for Further Maths students with no novel insight required. The telescoping pattern is straightforward once partial fractions are found, making this slightly easier than average.
Spec4.06b Method of differences: telescoping series
1 In this question you must show detailed reasoning.
  1. By using partial fractions show that \(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } + 3 r + 2 } = \frac { 1 } { 2 } - \frac { 1 } { n + 2 }\).
  2. Hence determine the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } + 3 r + 2 }\).
Question 1:
AnswerMarks Guidance
1(a) DR
(r + 2)(r + 1)
𝐴𝐴 𝐡𝐡
A = 1, B = –1 +
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+2
1 1 1 1 1 1 1
Ξ£= βˆ’ + βˆ’ + … βˆ’ +
2 3 3 4 4 𝑛𝑛+1 𝑛𝑛+1
1
1 1
= βˆ’ βˆ’
AnswerMarks
2 n+2 𝑛𝑛+2B1
M1
A1
M1
A1
AnswerMarks
[5]1.1a
1.1
1.1
2.1
AnswerMarks
1.1Correct factorisation of
denominator soi
Correct general form for partial
fractions soi by correct answer
Both
Condone omission of
for M1 only
1 1
+4 π‘Žπ‘Žπ‘›π‘›π‘Žπ‘Ž βˆ’π‘›π‘›+1
AnswerMarks
AG. Cancellation must be evidentCould be incorrect if recovered.
eg and C = 0
𝐴𝐴 𝐡𝐡
π‘Ÿπ‘Ÿ+1+π‘Ÿπ‘Ÿ+2+𝐢𝐢
1 1
βˆ’
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+2
AnswerMarks
(b)DR
∞ 1
βˆ‘ =
2
1
since β†’0 as nβ†’βˆž
AnswerMarks
n+2B1
B1
AnswerMarks
[2]2.2a
2.4Or
1 1 1 1
AnswerMarks
𝑛𝑛liβ†’m∞�2βˆ’π‘›π‘›+2οΏ½= 2βˆ’0 = 2Indication that 1/(n + 2) is close
to zero (accept β€œsmall”) when n is
large
Question 1:
1 | (a) | DR
(r + 2)(r + 1)
𝐴𝐴 𝐡𝐡
A = 1, B = –1 +
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+2
1 1 1 1 1 1 1
Ξ£= βˆ’ + βˆ’ + … βˆ’ +
2 3 3 4 4 𝑛𝑛+1 𝑛𝑛+1
1
1 1
= βˆ’ βˆ’
2 n+2 𝑛𝑛+2 | B1
M1
A1
M1
A1
[5] | 1.1a
1.1
1.1
2.1
1.1 | Correct factorisation of
denominator soi
Correct general form for partial
fractions soi by correct answer
Both
Condone omission of
for M1 only
1 1
+4 π‘Žπ‘Žπ‘›π‘›π‘Žπ‘Ž βˆ’π‘›π‘›+1
AG. Cancellation must be evident | Could be incorrect if recovered.
eg and C = 0
𝐴𝐴 𝐡𝐡
π‘Ÿπ‘Ÿ+1+π‘Ÿπ‘Ÿ+2+𝐢𝐢
1 1
βˆ’
π‘Ÿπ‘Ÿ+1 π‘Ÿπ‘Ÿ+2
(b) | DR
∞ 1
βˆ‘ =
2
1
since β†’0 as nβ†’βˆž
n+2 | B1
B1
[2] | 2.2a
2.4 | Or
1 1 1 1
𝑛𝑛liβ†’m∞�2βˆ’π‘›π‘›+2οΏ½= 2βˆ’0 = 2 | Indication that 1/(n + 2) is close
to zero (accept β€œsmall”) when n is
large
1 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item By using partial fractions show that $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } + 3 r + 2 } = \frac { 1 } { 2 } - \frac { 1 } { n + 2 }$.
\item Hence determine the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } + 3 r + 2 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2019 Q1 [7]}}
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Q1 7 Q2 8 Q3 5 Q4 5 Q5 11 Q6 6 Q7 7 Q8 8 Q9 11 Q10 7