OCR MEI Further Pure Core AS 2024 June — Question 3 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2024
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSums Between Limits
DifficultyStandard +0.3 This is a straightforward application of standard summation formulae requiring students to substitute 2n into the sum of cubes formula, then subtract the sum from 1 to n to find the sum from n+1 to 2n. The algebraic manipulation to factorize into the given form is routine. While it requires careful algebra, it involves no novel insight or problem-solving—just direct application of a standard technique taught explicitly in Further Maths courses.
Spec4.06a Summation formulae: sum of r, r^2, r^3
3
  1. Using standard summation formulae, write down an expression in terms of \(n\) for \(\sum _ { r = 1 } ^ { 2 n } r ^ { 3 }\).
  2. Hence show that \(\sum _ { \mathrm { r } = \mathrm { n } + 1 } ^ { 2 \mathrm { n } } \mathrm { r } ^ { 3 } = \frac { 1 } { 4 } \mathrm { n } ^ { 2 } ( \mathrm { an } + \mathrm { b } ) ( \mathrm { cn } + \mathrm { d } )\), where \(a , b , c\) and \(d\) are integers to be determined.
Question 3:
AnswerMarks Guidance
3(a) 2n
1
r3 = (2n)2(2n+1)2 [=n2(2n+1)2]
4
AnswerMarks Guidance
r=1B1
[1]1.1 1
oe, 2 n 2 ( 2 n + 1 ) 2 is B0, mark final answer
4
AnswerMarks Guidance
3(b) n 2 n 2r n
r 3 = r 3 − r r 3
r n = + 1 = 1 = 1
1
= n 2 ( 2 n + 1 ) 2 − n 2 ( n + 1 ) 2
4
1
= n 2 [ 4 ( 2 n + 1 ) 2 − ( n + 1 ) 2 ]
4
1
= n 2 ( 4 n + 2 + n + 1 ) ( 4 n + 2 − n − 1 )
4
1
= n 2 ( 5 n + 3 ) ( 3 n + 1 )
AnswerMarks
4M1
A1
M1
A1
A1
AnswerMarks
[5]2.5
1.1
2.1
2.1
AnswerMarks
2.2asum from n+1 to 2n is sum from 1 to 2n − sum from 1 to n
taking out common factor of n2 (at any stage)
1
or n 2 (1 5 n 2 + 1 4 n + 3 ) oe
4
Question 3:
3 | (a) | 2n
1
r3 = (2n)2(2n+1)2 [=n2(2n+1)2]
4
r=1 | B1
[1] | 1.1 | 1
oe, 2 n 2 ( 2 n + 1 ) 2 is B0, mark final answer
4
3 | (b) | n 2 n 2r n
r 3 = r 3 − r r 3
r n = + 1 = 1 = 1
1
= n 2 ( 2 n + 1 ) 2 − n 2 ( n + 1 ) 2
4
1
= n 2 [ 4 ( 2 n + 1 ) 2 − ( n + 1 ) 2 ]
4
1
= n 2 ( 4 n + 2 + n + 1 ) ( 4 n + 2 − n − 1 )
4
1
= n 2 ( 5 n + 3 ) ( 3 n + 1 )
4 | M1
A1
M1
A1
A1
[5] | 2.5
1.1
2.1
2.1
2.2a | sum from n+1 to 2n is sum from 1 to 2n − sum from 1 to n
taking out common factor of n2 (at any stage)
1
or n 2 (1 5 n 2 + 1 4 n + 3 ) oe
4
3
\begin{enumerate}[label=(\alph*)]
\item Using standard summation formulae, write down an expression in terms of $n$ for $\sum _ { r = 1 } ^ { 2 n } r ^ { 3 }$.
\item Hence show that $\sum _ { \mathrm { r } = \mathrm { n } + 1 } ^ { 2 \mathrm { n } } \mathrm { r } ^ { 3 } = \frac { 1 } { 4 } \mathrm { n } ^ { 2 } ( \mathrm { an } + \mathrm { b } ) ( \mathrm { cn } + \mathrm { d } )$, where $a , b , c$ and $d$ are integers to be determined.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2024 Q3 [6]}}
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