CAIE FP1 2017 November — Question 1 4 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2017
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSum of Powers Using Standard Formulae
DifficultyStandard +0.3 This is a straightforward application of standard summation formulae requiring expansion to 16r² - 8r - 3, then applying Σr² and Σr formulae. While it's Further Maths content, it's a routine algebraic manipulation with no conceptual challenges beyond remembering standard results, making it slightly easier than average.
Spec4.06a Summation formulae: sum of r, r^2, r^3
1 Find \(\sum _ { r = 1 } ^ { n } ( 4 r - 3 ) ( 4 r + 1 )\), giving your answer in its simplest form.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\sum_{r=1}^{n} u_r = 16\sum_{r=1}^{n} r^2 - 8\sum_{r=1}^{n} r - 3n\)M1A1 M1 for split into 3 parts
\(= 16\cdot\frac{n(n+1)(2n+1)}{6} - 8\cdot\frac{n(n+1)}{2} - 3n\)M1 For using formulae correctly in their expression
\(= \ldots = \frac{n}{3}(16n^2 + 12n - 13)\) (3 terms)A1 OE
Total4 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} u_r = 16\sum_{r=1}^{n} r^2 - 8\sum_{r=1}^{n} r - 3n$ | M1A1 | M1 for split into 3 parts |
| $= 16\cdot\frac{n(n+1)(2n+1)}{6} - 8\cdot\frac{n(n+1)}{2} - 3n$ | M1 | For using formulae correctly in their expression |
| $= \ldots = \frac{n}{3}(16n^2 + 12n - 13)$ (3 terms) | A1 | OE |
| **Total** | **4** | |

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1 Find $\sum _ { r = 1 } ^ { n } ( 4 r - 3 ) ( 4 r + 1 )$, giving your answer in its simplest form.\\

\hfill \mbox{\textit{CAIE FP1 2017 Q1 [4]}}
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Q1 4 Q2 6 Q3 7 Q4 8 Q5 8 Q6 9 Q7 10 Q8 11 Q9 12 Q10 12 Q11 EITHER Q11 OR