CAIE FP1 2014 June — Question 2 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeProving standard summation formulae
DifficultyStandard +0.8 This is a standard Further Maths proof using the method of differences to derive the sum of cubes formula. While it requires understanding telescoping sums and algebraic manipulation with given standard results, it's a well-established technique taught explicitly in FP1. The multi-step nature and need to connect several ideas places it above average, but it's not exceptionally challenging for Further Maths students who have practiced this method.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
2 Expand and simplify \(( r + 1 ) ^ { 4 } - r ^ { 4 }\). Use the method of differences together with the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$
Question 2:
AnswerMarks Guidance
Working/AnswerMark Guidance
\((r+1)^4 - r^4 = 4r^3 + 6r^2 + 4r + 1\)B1
\((n+1)^4 - 1^4 = 4\Sigma_{r=1}^{n} r^3 + 6\Sigma_{r=1}^{n} r^2 + 4\Sigma_{r=1}^{n} r + n\)M1
\(n^4 + 4n^3 + 6n^2 + 4n = 4\Sigma_{r=1}^{n} r^3 + n(2n^2 + 3n + 1) + 2n^2 + 2n + n\)A1A1
\(\Rightarrow \Sigma_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2\)A1 AG
Total: [4]
## Question 2:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $(r+1)^4 - r^4 = 4r^3 + 6r^2 + 4r + 1$ | B1 | |
| $(n+1)^4 - 1^4 = 4\Sigma_{r=1}^{n} r^3 + 6\Sigma_{r=1}^{n} r^2 + 4\Sigma_{r=1}^{n} r + n$ | M1 | |
| $n^4 + 4n^3 + 6n^2 + 4n = 4\Sigma_{r=1}^{n} r^3 + n(2n^2 + 3n + 1) + 2n^2 + 2n + n$ | A1A1 | |
| $\Rightarrow \Sigma_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2$ | A1 | AG |

**Total: [4]**

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2 Expand and simplify $( r + 1 ) ^ { 4 } - r ^ { 4 }$.

Use the method of differences together with the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that

$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$

\hfill \mbox{\textit{CAIE FP1 2014 Q2 [5]}}
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