OCR FP1 2007 June — Question 3 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2007
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeUse standard formulae to show result
DifficultyModerate -0.5 This is a straightforward algebraic manipulation question requiring recall of two standard summation formulae and basic algebra to verify the result. While it's from Further Maths FP1, it requires no proof technique, no induction despite the topic label, and no problem-solving—just direct substitution and simplification. It's easier than average but not trivial since it involves manipulating summation notation correctly.
Spec4.06a Summation formulae: sum of r, r^2, r^3

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

AnswerMarks Guidance
\(3\sum r^2 - 3\sum r + \sum 1\)M1 Consider the sum of three separate terms.
\(3\sum r^2 = \frac{1}{2}n(n+1)(2n+1)\)A1 Correct formula stated.
\(3\sum r = \frac{3}{2}n(n+1)\)A1 Correct formula stated.
\(\sum 1 = n\)A1 Correct term seen.
\(n^3\)M1 A1 Attempt to simplify. Obtain given answer correctly.
Total: 6 marks
$3\sum r^2 - 3\sum r + \sum 1$ | M1 | Consider the sum of three separate terms. |

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

$3\sum r = \frac{3}{2}n(n+1)$ | A1 | Correct formula stated. |

$\sum 1 = n$ | A1 | Correct term seen. |

$n^3$ | M1 A1 | Attempt to simplify. Obtain given answer correctly. |

**Total: 6 marks**
3 Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that, for all positive integers $n$,

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

\hfill \mbox{\textit{OCR FP1 2007 Q3 [6]}}