Question 5 - A-Level Maths

OCR MEI FP1 2013 June — Question 5 6 marks

Exam Board	OCR MEI
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and series, recurrence and convergence
Type	Method of differences with given identity
Difficulty	Standard +0.3 This is a straightforward application of the method of differences where the partial fraction identity is already provided. Students only need to write out terms, observe the telescoping pattern, and simplify—no derivation of the identity or complex algebraic manipulation required. Slightly easier than average due to the given identity.
Spec	1.02y Partial fractions: decompose rational functions 4.06b Method of differences: telescoping series

5 You are given that $\frac { 4 } { ( 4 n - 3 ) ( 4 n + 1 ) } \equiv \frac { 1 } { 4 n - 3 } - \frac { 1 } { 4 n + 1 }$. Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 4 r - 3 ) ( 4 r + 1 ) } = \frac { n } { 4 n + 1 }$$

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Question 5:

Answer	Marks	Guidance
$\sum_{r=1}^{n} \frac{1}{(4r-3)(4r+1)} = \frac{1}{4}\sum_{r=1}^{n}\left[\frac{1}{4r-3} - \frac{1}{4r+1}\right]$	M1	For splitting summation into two; allow missing $\frac{1}{4}$
$= \frac{1}{4}\left[\left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{9}\right) + \ldots + \left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)\right]$	M1	Write out terms (at least first and last terms in full)
A1	Allow missing $\frac{1}{4}$
$= \frac{1}{4}\left[1 - \frac{1}{4n+1}\right]$	M1	Cancelling inner terms; SC insufficient working shown above, M1M0M1A1 (allow missing $\frac{1}{4}$)
A1	Inclusion of $\frac{1}{4}$ justified
$= \frac{1}{4}\left[\frac{4n+1-1}{4n+1}\right] = \frac{n}{4n+1}$	A1	Honestly obtained (AG)

[6]

## Question 5:
$\sum_{r=1}^{n} \frac{1}{(4r-3)(4r+1)} = \frac{1}{4}\sum_{r=1}^{n}\left[\frac{1}{4r-3} - \frac{1}{4r+1}\right]$ | M1 | For splitting summation into two; allow missing $\frac{1}{4}$
$= \frac{1}{4}\left[\left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{9}\right) + \ldots + \left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)\right]$ | M1 | Write out terms (at least first and last terms in full)
| A1 | Allow missing $\frac{1}{4}$
$= \frac{1}{4}\left[1 - \frac{1}{4n+1}\right]$ | M1 | Cancelling inner terms; SC insufficient working shown above, M1M0M1A1 (allow missing $\frac{1}{4}$)
| A1 | Inclusion of $\frac{1}{4}$ justified
$= \frac{1}{4}\left[\frac{4n+1-1}{4n+1}\right] = \frac{n}{4n+1}$ | A1 | Honestly obtained (AG)
**[6]**

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5 You are given that $\frac { 4 } { ( 4 n - 3 ) ( 4 n + 1 ) } \equiv \frac { 1 } { 4 n - 3 } - \frac { 1 } { 4 n + 1 }$. Use the method of differences to show that

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 4 r - 3 ) ( 4 r + 1 ) } = \frac { n } { 4 n + 1 }$$

\hfill \mbox{\textit{OCR MEI FP1 2013 Q5 [6]}}

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Q1 5 Q2 6 Q3 6 Q4 6 Q5 6 Q6 7 Q7 12 Q8 12 Q9 12

Answer	Marks	Guidance
\(\sum_{r=1}^{n} \frac{1}{(4r-3)(4r+1)} = \frac{1}{4}\sum_{r=1}^{n}\left[\frac{1}{4r-3} - \frac{1}{4r+1}\right]\)	M1	For splitting summation into two; allow missing \(\frac{1}{4}\)
\(= \frac{1}{4}\left[\left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{9}\right) + \ldots + \left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)\right]\)	M1	Write out terms (at least first and last terms in full)
A1	Allow missing \(\frac{1}{4}\)
\(= \frac{1}{4}\left[1 - \frac{1}{4n+1}\right]\)	M1	Cancelling inner terms; SC insufficient working shown above, M1M0M1A1 (allow missing \(\frac{1}{4}\))
A1	Inclusion of \(\frac{1}{4}\) justified
\(= \frac{1}{4}\left[\frac{4n+1-1}{4n+1}\right] = \frac{n}{4n+1}\)	A1	Honestly obtained (AG)