CAIE FP1 2015 June — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve inequality: recurrence sequence
DifficultyChallenging +1.2 This is a two-part induction problem requiring students to prove an inequality holds for a recursively defined sequence, then prove the sequence is decreasing. While it involves Further Maths content (FP1), the inductive steps are relatively straightforward: the first part uses AM-GM inequality (4a_n/5 + 5/a_n ≥ 2√4 = 4, but need >5), and the second part requires algebraic manipulation to show a_n - a_{n+1} > 0. The techniques are standard for FP1 induction questions, though the recursive formula and two-part structure elevate it slightly above average difficulty.
Spec4.01a Mathematical induction: construct proofs
3 The sequence \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is such that \(a _ { 1 } > 5\) and \(a _ { n + 1 } = \frac { 4 a _ { n } } { 5 } + \frac { 5 } { a _ { n } }\) for every positive integer \(n\).
Prove by mathematical induction that \(a _ { n } > 5\) for every positive integer \(n\). Prove also that \(a _ { n } > a _ { n + 1 }\) for every positive integer \(n\).
Question 3:
AnswerMarks Guidance
WorkingMarks Notes
\(a_1 > 5\) (given) \(\Rightarrow H_1\) is trueB1
Assume \(H_k\) is true for some positive integer \(k\), i.e. \(a_k = 5 + \delta\), where \(\delta > 0\)B1
\(a_{k+1} - 5 = \frac{4a_k^2 + 25}{5a_k} - 5 = \frac{4a_k^2 + 25 - 25a_k}{5a_k} = \frac{(4a_k - 5)(a_k - 5)}{5a_k} > 0 \Rightarrow a_{k+1} > 5\)M1A1
Or: \(a_{k+1} = \frac{4}{5}(5+\delta) + \frac{5}{5+\delta} = 4 + \frac{4}{5}\delta + (1 - \frac{\delta}{5} + \frac{\delta^2}{25} - \ldots)\) for \(0 < \delta < 5\)(M1)
\(= 5 + \frac{3}{5}\delta + 0(\delta^2) \geq a_{k+1} > 5\), (\(\delta \geq 5\) is trivial)(A1)
\(H_k \Rightarrow H_{k+1}\) and \(H_1\) is true, hence by mathematical induction result true for all \(n \in \mathbb{Z}^+\)A1 (5) N.B. minimum: 'true for all positive integers'
\(a_{k+1} - a_k = \frac{5}{a_k} - \frac{1}{5}a_k\)M1
\(\frac{5}{a_k} < 1\) and \(\frac{1}{5}a_k > 1 \Rightarrow a_{k+1} - a_k < 0 \Rightarrow a_{k+1} < a_k\)A1 (2)
Total: 7 
## Question 3:

| Working | Marks | Notes |
|---------|-------|-------|
| $a_1 > 5$ (given) $\Rightarrow H_1$ is true | B1 | |
| Assume $H_k$ is true for some positive integer $k$, i.e. $a_k = 5 + \delta$, where $\delta > 0$ | B1 | |
| $a_{k+1} - 5 = \frac{4a_k^2 + 25}{5a_k} - 5 = \frac{4a_k^2 + 25 - 25a_k}{5a_k} = \frac{(4a_k - 5)(a_k - 5)}{5a_k} > 0 \Rightarrow a_{k+1} > 5$ | M1A1 | |
| **Or:** $a_{k+1} = \frac{4}{5}(5+\delta) + \frac{5}{5+\delta} = 4 + \frac{4}{5}\delta + (1 - \frac{\delta}{5} + \frac{\delta^2}{25} - \ldots)$ for $0 < \delta < 5$ | (M1) | |
| $= 5 + \frac{3}{5}\delta + 0(\delta^2) \geq a_{k+1} > 5$, ($\delta \geq 5$ is trivial) | (A1) | |
| $H_k \Rightarrow H_{k+1}$ and $H_1$ is true, hence by mathematical induction result true for all $n \in \mathbb{Z}^+$ | A1 | (5) N.B. minimum: 'true for all positive integers' |
| $a_{k+1} - a_k = \frac{5}{a_k} - \frac{1}{5}a_k$ | M1 | |
| $\frac{5}{a_k} < 1$ and $\frac{1}{5}a_k > 1 \Rightarrow a_{k+1} - a_k < 0 \Rightarrow a_{k+1} < a_k$ | A1 | (2) |
| | **Total: 7** | |

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3 The sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is such that $a _ { 1 } > 5$ and $a _ { n + 1 } = \frac { 4 a _ { n } } { 5 } + \frac { 5 } { a _ { n } }$ for every positive integer $n$.\\
Prove by mathematical induction that $a _ { n } > 5$ for every positive integer $n$.

Prove also that $a _ { n } > a _ { n + 1 }$ for every positive integer $n$.

\hfill \mbox{\textit{CAIE FP1 2015 Q3 [7]}}
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