CAIE FP1 2013 November — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 This question involves standard manipulation of series notation and the relationship between partial sums and individual terms. Finding u_r from S_n uses the formula u_r = S_r - S_{r-1}, which is a routine technique. The final part requires recognizing that the sum from n+1 to 2n equals S_{2n} - S_n, then substituting and simplifying polynomials. While it requires multiple steps and careful algebraic manipulation, all techniques are standard for Further Maths students and no novel insight is needed.
Spec4.06a Summation formulae: sum of r, r^2, r^3
3 It is given that $$S _ { n } = \sum _ { r = 1 } ^ { n } u _ { r } = 2 n ^ { 2 } + n$$ Write down the values of \(S _ { 1 } , S _ { 2 } , S _ { 3 } , S _ { 4 }\). Express \(u _ { r }\) in terms of \(r\), justifying your answer. Find $$\sum _ { r = n + 1 } ^ { 2 n } u _ { r } .$$
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S_1\ldots S_4 \sim 3, 10, 21, 36\)B1 Writes first four sums
\(u_1\ldots u_4 \sim 3, 7, 11, 15 \Rightarrow u_r = 4r-1\)B1B1 Deduces first four terms, conjectures
since \(S_n = \frac{n}{2}\{6+4(n-1)\} = 2n^2+n\) as givenB1 Justifies result
Or \(u_r = S_r - S_{r-1} = 2r^2+r-2(r-1)^2-(r-1) = 4r-1\)B1B1 Alternative method
\(\sum_{n+1}^{2n}(4r-1) = 4\cdot\frac{2n(2n+1)}{2} - 2n - \left(4\cdot\frac{n(n+1)}{2}-n\right)\)M1A1 Obtains required sum
\(= 8n^2+2n-(2n^2+n) = 6n^2+n\)A1
Or Sum of AP \(= \frac{n}{2}(4n+3+8n-1) = 6n^2+n\) Alternative 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_1\ldots S_4 \sim 3, 10, 21, 36$ | B1 | Writes first four sums |
| $u_1\ldots u_4 \sim 3, 7, 11, 15 \Rightarrow u_r = 4r-1$ | B1B1 | Deduces first four terms, conjectures |
| since $S_n = \frac{n}{2}\{6+4(n-1)\} = 2n^2+n$ as given | B1 | Justifies result |
| **Or** $u_r = S_r - S_{r-1} = 2r^2+r-2(r-1)^2-(r-1) = 4r-1$ | B1B1 | Alternative method |
| $\sum_{n+1}^{2n}(4r-1) = 4\cdot\frac{2n(2n+1)}{2} - 2n - \left(4\cdot\frac{n(n+1)}{2}-n\right)$ | M1A1 | Obtains required sum |
| $= 8n^2+2n-(2n^2+n) = 6n^2+n$ | A1 | |
| Or Sum of AP $= \frac{n}{2}(4n+3+8n-1) = 6n^2+n$ | | Alternative |

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3 It is given that

$$S _ { n } = \sum _ { r = 1 } ^ { n } u _ { r } = 2 n ^ { 2 } + n$$

Write down the values of $S _ { 1 } , S _ { 2 } , S _ { 3 } , S _ { 4 }$. Express $u _ { r }$ in terms of $r$, justifying your answer.

Find

$$\sum _ { r = n + 1 } ^ { 2 n } u _ { r } .$$

\hfill \mbox{\textit{CAIE FP1 2013 Q3 [7]}}
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