CAIE FP1 2016 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2016
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeInfinite series convergence and sum
DifficultyStandard +0.8 This question requires partial fraction decomposition of 1/((2r)²-1), applying the method of differences to find telescoping cancellation, and then taking a limit as n→∞. While method of differences is a standard Further Maths technique, students must recognize the factorization (2r)²-1 = (2r-1)(2r+1), set up the correct partial fractions, identify the telescoping pattern, and handle the limit carefully. This is moderately challenging for FP1 level, requiring multiple connected steps and careful algebraic manipulation.
Spec4.06b Method of differences: telescoping series
1 Use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }\). Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)\)M1A1
\(\sum_{r=1}^{n} \frac{1}{(2r)^2-1} = \frac{1}{2}\left(\left[1-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{5}\right]+\ldots+\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]\right) = \frac{1}{2}\left(1-\frac{1}{2n+1}\right)\)M1A1 OE, [4]
\(\frac{1}{2}\left(1-\frac{1}{2n+1}\right) = \frac{n}{2n+1} \Rightarrow \sum_{r=1}^{\infty}\frac{1}{(2r)^2-1} = \frac{1}{2}\)B1\(\checkmark\) [1] 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)$ | M1A1 | |
| $\sum_{r=1}^{n} \frac{1}{(2r)^2-1} = \frac{1}{2}\left(\left[1-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{5}\right]+\ldots+\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]\right) = \frac{1}{2}\left(1-\frac{1}{2n+1}\right)$ | M1A1 | OE, [4] |
| $\frac{1}{2}\left(1-\frac{1}{2n+1}\right) = \frac{n}{2n+1} \Rightarrow \sum_{r=1}^{\infty}\frac{1}{(2r)^2-1} = \frac{1}{2}$ | B1$\checkmark$ | [1] |

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1 Use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }$.

Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }$.

\hfill \mbox{\textit{CAIE FP1 2016 Q1 [5]}}
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