Question 5 - A-Level Maths

OCR FP1 2016 June — Question 5 4 marks

Exam Board	OCR
Module	FP1 (Further Pure Mathematics 1)
Year	2016
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove recurrence relation formula
Difficulty	Standard +0.3 This is a straightforward proof by induction of a recurrence relation formula. The base case is trivial (u₁ = 5 checks out), and the inductive step requires only substituting the assumed formula into the recurrence relation and simplifying algebraically. While it's a Further Maths topic, it follows a completely standard template with no conceptual challenges or novel insights required, making it slightly easier than average overall.
Spec	4.01a Mathematical induction: construct proofs

5 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by $$u _ { 1 } = 5 \text { and } u _ { n + 1 } = 3 u _ { n } + 2 \text { for } n \geqslant 1 \text {. }$$ Prove by induction that $u _ { n } = 2 \times 3 ^ { n } - 1$.

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Question 5:

Answer	Marks	Guidance
Answer	Marks	Guidance
$3(2\times3^n - 1)+2$	B1	Show clearly result true for $n=1$, accept $(u_1)=2\times3-1=5$
M1	Substitute for $u_n$ in recurrence relation
A1	Establish correct result for $u_{n+1}$ convincingly
B1	Clear statement of induction conclusion, provided first 3 marks earned
[4]

## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3(2\times3^n - 1)+2$ | B1 | Show clearly result true for $n=1$, accept $(u_1)=2\times3-1=5$ |
| | M1 | Substitute for $u_n$ in recurrence relation |
| | A1 | Establish correct result for $u_{n+1}$ convincingly |
| | B1 | Clear statement of induction conclusion, provided first 3 marks earned |
| **[4]** | | |

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5 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by

$$u _ { 1 } = 5 \text { and } u _ { n + 1 } = 3 u _ { n } + 2 \text { for } n \geqslant 1 \text {. }$$

Prove by induction that $u _ { n } = 2 \times 3 ^ { n } - 1$.

\hfill \mbox{\textit{OCR FP1 2016 Q5 [4]}}

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