OCR MEI Further Pure Core 2019 June — Question 1 4 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyEasy -1.2 This is a straightforward application of standard summation formulae requiring splitting the sum, applying ∑r² = n(n+1)(2n+1)/6, subtracting n, and factorising. It's routine bookwork with no problem-solving element, though the factorisation adds a small step beyond pure recall.
Spec4.06a Summation formulae: sum of r, r^2, r^3

1 Find \(\sum _ { r = 1 } ^ { n } \left( 2 r ^ { 2 } - 1 \right)\), expressing your answer in fully factorised form.

Question 1:
AnswerMarks
1n 1
 (2r2 1) n(n1)(2n1)n
3
r1
1
 n(2n23n2)
3
1
 n(2n1)(n2)
AnswerMarks
3B1
B1
B1
B1cao
AnswerMarks
[4]2.5
1.1b
1.1b
AnswerMarks
1.1b1
n(n1)(2n1)...
3
… –n
factoring out n
2
allow n(2n1)(n2), etc
AnswerMarks
6correctly
Question 1:
1 | n 1
 (2r2 1) n(n1)(2n1)n
3
r1
1
 n(2n23n2)
3
1
 n(2n1)(n2)
3 | B1
B1
B1
B1cao
[4] | 2.5
1.1b
1.1b
1.1b | 1
n(n1)(2n1)...
3
… –n
factoring out n
2
allow n(2n1)(n2), etc
6 | correctly
1 Find $\sum _ { r = 1 } ^ { n } \left( 2 r ^ { 2 } - 1 \right)$, expressing your answer in fully factorised form.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q1 [4]}}