OCR FP1 2006 June — Question 4 5 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.5 This is a straightforward application of standard summation formulae (∑r³ and ∑r²) requiring only algebraic manipulation to combine them and factorize the result. While it's Further Maths content, it's a routine bookwork-style question with no problem-solving or novel insight required, making it easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3

4 Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + r ^ { 2 } \right) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + 1 )$$

AnswerMarks Guidance
\(\Sigma r^3 + \Sigma r^2\)M1 Consider the sum as two separate parts
\(\Sigma r^2 = \frac{1}{6}n(n+1)(2n+1)\)A1 Correct formula stated
\(\Sigma r^3 = \frac{1}{4}n^2(n+1)^2\)A1 Correct formula stated
\(\frac{1}{12}n(n+1)(n+2)(3n+1)\)M1 Attempt to factorise and simplify or expand both expressions
A1, 5Obtain given answer correctly or complete verification
$\Sigma r^3 + \Sigma r^2$ | M1 | Consider the sum as two separate parts

$\Sigma r^2 = \frac{1}{6}n(n+1)(2n+1)$ | A1 | Correct formula stated

$\Sigma r^3 = \frac{1}{4}n^2(n+1)^2$ | A1 | Correct formula stated

$\frac{1}{12}n(n+1)(n+2)(3n+1)$ | M1 | Attempt to factorise and simplify or expand both expressions
| A1, 5 | Obtain given answer correctly or complete verification
4 Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that, for all positive integers $n$,

$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + r ^ { 2 } \right) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + 1 )$$

\hfill \mbox{\textit{OCR FP1 2006 Q4 [5]}}