OCR MEI Further Pure Core AS 2018 June — Question 7 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFinding n for given sum value
DifficultyStandard +0.8 This is a telescoping series problem requiring partial fractions, series summation, and solving an inequality. Part (i) is routine (2 marks), but part (ii) requires recognizing the telescoping pattern, deriving the sum formula S_n = 1 - 1/(2n+1), then solving 1 - 1/(2n+1) > 0.999999 to find n > 499999.5. While methodical, this multi-step problem with large numbers and careful algebraic manipulation is moderately challenging for AS Further Maths.
Spec4.06b Method of differences: telescoping series
  1. Express \(\frac{1}{2r-1} - \frac{1}{2r+1}\) as a single fraction. [2]
  2. Find how many terms of the series $$\frac{2}{1 \times 3} + \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \ldots + \frac{2}{(2r-1)(2r+1)} + \ldots$$ are needed for the sum to exceed \(0.999999\). [7]
Question 7:
AnswerMarks Guidance
7(i) 1 1 2r1(2r1) 2
  
AnswerMarks
2r1 2r1 (2r1)(2r1) (2r1)(2r1)M1
A1
AnswerMarks
[2]1.1a
1.1combining fractions
2
or
AnswerMarks
4r2 1mark final answer
M1
A1
M1
A1
(z – 2 + i)(z  2 – i)
= z2 – 4z + 5
their (z2  4z + 5)  (z  3)
(z – 2 + i)(z  2 – i) = z2 – 4z + 5
(z2  4z + 5)(z  3) = z3  7z2 + 17z  15
so eqn is z3  7z2 + 17z  15 = 0
AnswerMarks Guidance
7(ii) 2 2 2 2
  K 
13 35 57 (2n1)(2n1)
n 2

(2r1)(2r1)
r1
n  1 1 
 
2r1 2r1
r1
1 1 1 1 1 1 1
1     K  
3 3 5 5 7 2n1 2n1
1
1
2n1
1
so 1 0.999999
2n1
 2n + 1 > 1000000  n > 499999.5
AnswerMarks
 Number of terms is 500000M1
M1
M1
A1
M1
M1
A1cao
AnswerMarks
[7]3.1a
3.1a
2.1
2.1
1.1a
1.1
AnswerMarks
3.2a2
rth term is soi
(2r1)(2r1)
splitting into partial fractions soi
1
must end with ...
2n1
1
1 is A0
2r1
2n
or 0.999999 allow
2n1
equality, condone r
AnswerMarks
re-arranging (correctly) for n1
not ...
2r1
dep use of difference
method
[e.g. not by inspection]
trial and error: must show
both n = 499999 and
n = 500000 for final A1
Question 7:
7 | (i) | 1 1 2r1(2r1) 2
  
2r1 2r1 (2r1)(2r1) (2r1)(2r1) | M1
A1
[2] | 1.1a
1.1 | combining fractions
2
or
4r2 1 | mark final answer
M1
A1
M1
A1
(z – 2 + i)(z  2 – i)
= z2 – 4z + 5
their (z2  4z + 5)  (z  3)
(z – 2 + i)(z  2 – i) = z2 – 4z + 5
(z2  4z + 5)(z  3) = z3  7z2 + 17z  15
so eqn is z3  7z2 + 17z  15 = 0
7 | (ii) | 2 2 2 2
  K 
13 35 57 (2n1)(2n1)
n 2

(2r1)(2r1)
r1
n  1 1 
 
2r1 2r1
r1
1 1 1 1 1 1 1
1     K  
3 3 5 5 7 2n1 2n1
1
1
2n1
1
so 1 0.999999
2n1
 2n + 1 > 1000000  n > 499999.5
 Number of terms is 500000 | M1
M1
M1
A1
M1
M1
A1cao
[7] | 3.1a
3.1a
2.1
2.1
1.1a
1.1
3.2a | 2
rth term is soi
(2r1)(2r1)
splitting into partial fractions soi
1
must end with ...
2n1
1
1 is A0
2r1
2n
or 0.999999 allow
2n1
equality, condone r
re-arranging (correctly) for n | 1
not ...
2r1
dep use of difference
method
[e.g. not by inspection]
trial and error: must show
both n = 499999 and
n = 500000 for final A1
\begin{enumerate}[label=(\roman*)]
\item Express $\frac{1}{2r-1} - \frac{1}{2r+1}$ as a single fraction. [2]
\item Find how many terms of the series
$$\frac{2}{1 \times 3} + \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \ldots + \frac{2}{(2r-1)(2r+1)} + \ldots$$
are needed for the sum to exceed $0.999999$. [7]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2018 Q7 [9]}}
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