Recurrence Relations: Linear recurrence: behavior analysis — RVHS 2025 H2 Math Prelim Paper 2
What this question tests
Question
A sequence is such that \(u_1 = N\) where \(N\) is a constant, and \(u_{n+1} = 3u_n - 8\), for \(n \geq 1\).
- Describe how the sequence behaves when
- \(N = 4\),
- \(N = 3\).
- Find the value of \(N\) for which \(u_5 = 409\).
- The sequence \(v_n = (-1)^n(u_n + k)\) for \(n \geq 1\) where \(k\) is a constant, is convergent when \(u_1 = 4\). State the value of \(k\).
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(a)(i) For \(N = 4\), the sequence takes on a constant value with \(u_n = 4\) for all \(n \geq 1\).
(a)(ii) For \(N = 3\): \[\begin{aligned} u_1 &= 3\\ u_2 &= 3(3) - 8 = 1\\ u_3 &= 3(-5) - 8 = -23\\ &\vdots \end{aligned}\]
The sequence decreases and approaches negative infinity.
(b)
Method 1 — Working backwards from \(u_5 = 409\)
Using \(u_{n+1} = 3u_n - 8\): \[\begin{aligned} 3u_4 - 8 = 409 &\implies u_4 = 139\\ 3u_3 - 8 = 139 &\implies u_3 = 49\\ 3u_2 - 8 = 49 &\implies u_2 = 19\\ 3u_1 - 8 = 19 &\implies u_1 = 9 = N \end{aligned}\]
Method 2 — Express \(u_n\) in terms of \(N\)
\[\begin{aligned} u_1 &= N\\ u_2 &= 3N - 8\\ u_3 &= 9N - 32\\ u_4 &= 27N - 104\\ u_5 &= 81N - 320 \end{aligned}\] \[ 81N - 320 = 409 \implies N = 9 \](c) From part (a), when \(u_1 = 4\), \(u_n = 4\) for \(n \geq 1\).
\[ v_n = (-1)^n(u_n + k) \] \[\begin{aligned} v_1 &= -(4 + k)\\ v_2 &= +(4 + k)\\ v_3 &= -(4 + k)\\ &\vdots \end{aligned}\]For \(v_n = (-1)^n(u_n + k)\) to be convergent: