CAIE FP1 2012 June — Question 2 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyStandard +0.3 This is a straightforward induction proof with a given formula to verify. The recurrence relation substitution is algebraic but routine, requiring only fraction manipulation and simplification. While it's a Further Maths topic, the mechanics are standard: verify base case, assume for n=k, prove for n=k+1 by substituting into the recurrence relation. No novel insight or complex algebraic manipulation required beyond typical A-level techniques.
Spec4.01a Mathematical induction: construct proofs

2 For the sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\), it is given that \(u _ { 1 } = 1\) and \(u _ { r + 1 } = \frac { 3 u _ { r } - 2 } { 4 }\) for all \(r\). Prove by mathematical induction that \(u _ { n } = 4 \left( \frac { 3 } { 4 } \right) ^ { n } - 2\), for all positive integers \(n\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((P_n: u_n = 4\left(\frac{3}{4}\right)^n - 2)\) States proposition
Let \(n=1\): \(4 \times \frac{3}{4} - 2 = 3-2 = 1 \Rightarrow P_1\) trueB1 Proves base case
Assume \(P_k\) is true for some \(k\)B1 States inductive hypothesis
\(u_{k+1} = \frac{3\left\{4\left(\frac{3}{4}\right)^k-2\right\}-2}{4} = 4\cdot\frac{3}{4}\left(\frac{3}{4}\right)^k - \frac{6+2}{4}\)M1 Proves inductive step
\(= 4\left(\frac{3}{4}\right)^{k+1} - 2 \therefore P_k \Rightarrow P_{k+1}\)A1
\(\therefore\) By PMI \(P_n\) is true \(\forall\) positive integersA1 States conclusion; Part Mark: 5
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(P_n: u_n = 4\left(\frac{3}{4}\right)^n - 2)$ | — | States proposition |
| Let $n=1$: $4 \times \frac{3}{4} - 2 = 3-2 = 1 \Rightarrow P_1$ true | B1 | Proves base case |
| Assume $P_k$ is true for some $k$ | B1 | States inductive hypothesis |
| $u_{k+1} = \frac{3\left\{4\left(\frac{3}{4}\right)^k-2\right\}-2}{4} = 4\cdot\frac{3}{4}\left(\frac{3}{4}\right)^k - \frac{6+2}{4}$ | M1 | Proves inductive step |
| $= 4\left(\frac{3}{4}\right)^{k+1} - 2 \therefore P_k \Rightarrow P_{k+1}$ | A1 | |
| $\therefore$ By PMI $P_n$ is true $\forall$ positive integers | A1 | States conclusion; Part Mark: 5 |

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2 For the sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$, it is given that $u _ { 1 } = 1$ and $u _ { r + 1 } = \frac { 3 u _ { r } - 2 } { 4 }$ for all $r$. Prove by mathematical induction that $u _ { n } = 4 \left( \frac { 3 } { 4 } \right) ^ { n } - 2$, for all positive integers $n$.

\hfill \mbox{\textit{CAIE FP1 2012 Q2 [5]}}