CAIE FP1 2014 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSurd rationalization method of differences
DifficultyStandard +0.8 This is a telescoping series question requiring recognition of the method of differences pattern, careful tracking of which terms survive after cancellation (starting from k=13, not k=1), and evaluation of a limit as nā†’āˆž. While the algebraic manipulation is straightforward once the pattern is identified, the non-standard starting index and the need to handle surds in the limit requires more careful reasoning than typical method of differences questions that start at k=1.
Spec4.06b Method of differences: telescoping series
1 Given that $$u _ { k } = \frac { 1 } { \sqrt { } ( 2 k - 1 ) } - \frac { 1 } { \sqrt { } ( 2 k + 1 ) }$$ express \(\sum _ { k = 13 } ^ { n } u _ { k }\) in terms of \(n\). Deduce the value of \(\sum _ { k = 13 } ^ { \infty } u _ { k }\).
Question 1:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\ldots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right)\)M1A1 Telescoping series identified
\(\sum_{r=13}^{n} u_k = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\)M1A1 (4)
\(\sum_{r=13}^{\infty} u_k = \frac{1}{5}\)B1\(\checkmark\) (1) Follow through from previous result 
## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\ldots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right)$ | M1A1 | Telescoping series identified |
| $\sum_{r=13}^{n} u_k = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}$ | M1A1 (4) | |
| $\sum_{r=13}^{\infty} u_k = \frac{1}{5}$ | B1$\checkmark$ (1) | Follow through from previous result |

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1 Given that

$$u _ { k } = \frac { 1 } { \sqrt { } ( 2 k - 1 ) } - \frac { 1 } { \sqrt { } ( 2 k + 1 ) }$$

express $\sum _ { k = 13 } ^ { n } u _ { k }$ in terms of $n$.

Deduce the value of $\sum _ { k = 13 } ^ { \infty } u _ { k }$.

\hfill \mbox{\textit{CAIE FP1 2014 Q1 [5]}}
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