OCR Further Pure Core 1 2023 June — Question 1 3 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2023
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyStandard +0.3 This is a straightforward application of standard summation formulae requiring splitting the sum into two parts (Σr² and Σr³) and substituting known formulae. While it's a Further Maths question, it's a routine textbook exercise with no conceptual difficulty beyond formula manipulation, making it slightly easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3
1 In this question you must show detailed reasoning.
Determine the value of \(\sum _ { r = 1 } ^ { 50 } r ^ { 2 } ( 16 - r )\).
Question 1:
AnswerMarks
1DR
5 0 5 0 5 0
 r ( 2 1 6 − r ) =  1 6 r 2 −  r 3
1 1 1
1 6 1
=  5 0  5 1  1 0 1 − 5 0 2  5 1 2
6 4)
( = 6 8 6 8 0 0 − 1 6 2 5 6 2 5
AnswerMarks
= − 9 3 8 8 2 5M1
M1
AnswerMarks
A1Separating and using correct formulae for  r3 and  r2
Substituting must be seen
Including all notation correct (Sigmas do not need limits)
AnswerMarks
Alternative method:M1
M1
AnswerMarks
A1Separating and using the correct formulae
Substituting anywhere in the algebra
n 5 0 5 0
 r ( 2 1 6 − r )  2 = 1 6 r −  3 r
1 1 1
1 6 1
= ( n n + 1 ) + − ( 2 1 ) n n 2 ( n + 1 2 )
6 4
 8 1 
= ( n + n 1 )  + − ( 2 1 ) n n ( n + 1 )
3  4
( n + n 1 ) ( )
= 3 + − 2 6 1 3 n n 2
1 2
5 0  5 1 ( )
= 3 2  + − 6 1 5 0 3  5 2 0
1 2
= − 9 3 8 8 2 5
[3]
M1
M1
A1
Separating and using the correct formulae
Substituting anywhere in the algebra
Question 1:
1 | DR
5 0 5 0 5 0
 r ( 2 1 6 − r ) =  1 6 r 2 −  r 3
1 1 1
1 6 1
=  5 0  5 1  1 0 1 − 5 0 2  5 1 2
6 4)
( = 6 8 6 8 0 0 − 1 6 2 5 6 2 5
= − 9 3 8 8 2 5 | M1
M1
A1 | Separating and using correct formulae for  r3 and  r2
Substituting must be seen
Including all notation correct (Sigmas do not need limits)
Alternative method: | M1
M1
A1 | Separating and using the correct formulae
Substituting anywhere in the algebra
n 5 0 5 0
 r ( 2 1 6 − r )  2 = 1 6 r −  3 r
1 1 1
1 6 1
= ( n n + 1 ) + − ( 2 1 ) n n 2 ( n + 1 2 )
6 4
 8 1 
= ( n + n 1 )  + − ( 2 1 ) n n ( n + 1 )
3  4
( n + n 1 ) ( )
= 3 + − 2 6 1 3 n n 2
1 2
5 0  5 1 ( )
= 3 2  + − 6 1 5 0 3  5 2 0
1 2
= − 9 3 8 8 2 5
[3]
M1
M1
A1
Separating and using the correct formulae
Substituting anywhere in the algebra
1 In this question you must show detailed reasoning.\\
Determine the value of $\sum _ { r = 1 } ^ { 50 } r ^ { 2 } ( 16 - r )$.

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q1 [3]}}
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Q1 3 Q2 6 Q3 5 Q4 11 Q5 6 Q6 4 Q7 11 Q8 15 Q9 14