Recurrence Relations: Linear recurrence: limit & GP series sum — SAJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
A sequence \(u_1, u_2, u_3, \ldots\) is such that \(23u_{r+1} = 19u_r + 16\), for \(r>0\) and \(u_1 = 2\).
- Write down the value of \(u_{10}\), giving your answer correct to 4 decimal places.
- It is given that as \(r\to\infty\), \(u_r \to l\). Show that \(l=4\).
- Hence, find the smallest integer \(r\) for which \(u_r\) exceeds 99.9% of \(l\).
It is known that the \(k\)th term of this sequence is given by \(u_k = 4 - 2\!\left(\dfrac{19}{23}\right)^{\!k-1}\).
- Given that a new series \(S_n\) is defined by \(S_n = \displaystyle\sum_{k=1}^{n}(u_k - c)\), where \(c\) is a real constant.
Express \(S_n\) in the form \(An + B\!\left[1-\!\left(\dfrac{19}{23}\right)^{\!n}\right]\), where \(A\) is a constant in terms of \(c\) and \(B\) is a real number.
- Determine the possible values of \(c\) for \(S_n\) to diverge as \(n\to\infty\).
Show full worked solution▾
(i) From \(23u_{r+1} = 19u_r + 16\): \[ u_{r+1} = \frac{19}{23}u_r + \frac{16}{23} \]
From GC, \(u_{10} = 3.6417\).
(ii) As \(r\to\infty\), \(u_r \to l\) and \(u_{r+1} \to l\): \[ 23l = 19l + 16 \implies 4l = 16 \implies l = 4 \]
(iii) When \(u_r\) exceeds 99.9% of \(l\): \[ u_r > \frac{99.9}{100}(4) = 3.996 \]
Using \(u_{r+1} = \dfrac{19}{23}u_r + \dfrac{16}{23}\) on GC:
| \(r\) | \(u_r\) |
|---|---|
| 33 | \(3.9956 < 3.996\) |
| 34 | \(3.9963 > 3.996\) |
(iv) From \(u_k = 4 - 2\!\left(\dfrac{19}{23}\right)^{\!k-1}\): \[\begin{aligned} S_n &= \sum_{k=1}^{n}(u_k - c) = \sum_{k=1}^{n}\left[(4-c) - 2\left(\frac{19}{23}\right)^{\!k-1}\right]\\ &= (4-c)n - 2\cdot\frac{1-\left(\frac{19}{23}\right)^n}{1-\frac{19}{23}}\\ &= (4-c)n - \frac{23}{2}\!\left[1 - \left(\frac{19}{23}\right)^{\!n}\right] \end{aligned}\] so \(A = 4-c\) and \(B = -\dfrac{23}{2}\).
(v) As \(n\to\infty\), \(\left(\dfrac{19}{23}\right)^n\to 0\), so \(S_n\) diverges iff the linear term \((4-c)n\) does not vanish, i.e. \(c \neq 4\).