CAIE FP1 2011 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeTwo linear factors in denominator
DifficultyStandard +0.3 This is a standard Further Maths partial fractions question with method of differences. The partial fractions decomposition is straightforward with two linear factors, the telescoping series is routine to identify and sum, and finding the limit as n→∞ requires only basic understanding. While it's a multi-part question requiring several techniques, each step follows a well-practiced procedure with no novel insight needed. Slightly easier than average due to its predictable structure.
Spec1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series
1 Express \(\frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }\) in partial fractions and hence use the method of differences to find $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$ Deduce the value of $$\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{(2r+1)(2r+3)} = \frac{1}{2}\left(\frac{1}{2r+1} - \frac{1}{2r+3}\right)\)B1 Any method including cover-up rule
\(S_n = \frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)\cdots\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\right]\)M1A1 Expresses all terms as differences
\(= \frac{1}{6} - \frac{1}{2(2n+3)}\) (acf)A1 Part mark: 4
\(S_\infty = \frac{1}{6}\)A1 B0M1A1√ A0A1√ if signs reversed; Part mark: 1 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(2r+1)(2r+3)} = \frac{1}{2}\left(\frac{1}{2r+1} - \frac{1}{2r+3}\right)$ | B1 | Any method including cover-up rule |
| $S_n = \frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)\cdots\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\right]$ | M1A1 | Expresses all terms as differences |
| $= \frac{1}{6} - \frac{1}{2(2n+3)}$ (acf) | A1 | Part mark: 4 |
| $S_\infty = \frac{1}{6}$ | A1 | B0M1A1√ A0A1√ if signs reversed; Part mark: 1 |

---
1 Express $\frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$ in partial fractions and hence use the method of differences to find

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$

Deduce the value of

$$\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$

\hfill \mbox{\textit{CAIE FP1 2011 Q1 [5]}}
This paper (12 questions)
View full paper
Q1 5 Q2 6 Q3 6 Q4 6 Q5 8 Q6 9 Q7 11 Q8 11 Q9 12 Q10 12 Q11 EITHER Q11 OR