Standard +0.3 This is a straightforward proof by induction with a simple recurrence relation. The base case is trivial (u₁ = 2), and the inductive step requires only basic algebraic manipulation of the recurrence relation u_{n+1} = 2u_n - 1. While it's a Further Maths question, it's a standard textbook exercise testing the mechanical application of the induction framework rather than requiring any insight or problem-solving.
3 The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by \(u _ { 1 } = 2\), and \(u _ { n + 1 } = 2 u _ { n } - 1\) for \(n \geqslant 1\). Prove by induction that \(u _ { n } = 2 ^ { n - 1 } + 1\).
Use given result in recurrence relation in a relevant way
A1*
Obtain \(2^n + 1\) correctly
depA1 4
Specific statement of induction conclusion
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1* | Establish result true for $n = 1$ or $2$ |
| | M1* | Use given result in recurrence relation in a relevant way |
| | A1* | Obtain $2^n + 1$ correctly |
| | depA1 **4** | Specific statement of induction conclusion |
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