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Recurrence Relations: Periodic sequence & sigma notation — ACJC 2025 H2 Math Prelim Paper 2

Anglo-Chinese Junior College2025 PrelimPaper 2●●○ Standard9 marks

What this question tests

Periodic sequence & sigma notation.

Question

  1. For this question, you may use these results: \[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \qquad\text{and}\qquad \sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}. \]
    1. Show that \(\displaystyle\sum_{r=1}^{n} r(r+1)^2 = \frac{n(n+1)(n+2)(3n+5)}{12}\).
    2. Hence find \(\displaystyle\sum_{r=5}^{n-1}(r+2)(r+3)^2\) in terms of \(n\).
  2. The sequence \(u_1, u_2, u_3, \ldots\) is defined by \(u_1 = 2\), \(u_{n+1} = \dfrac{1}{1 - u_n}\), \(n \geq 1\).

    Find the values of \(u_2\), \(u_3\) and \(u_4\). Hence find the value of \(u_{2025}\).

Show full worked solution▾

Part (a)(i)

\[\begin{aligned} \sum_{r=1}^{n} r(r+1)^2 &= \sum_{r=1}^{n}\left(r^3 + 2r^2 + r\right)\\ &= \frac{n^2(n+1)^2}{4} + 2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\\ &= \frac{3n^2(n+1)^2}{12} + \frac{4n(n+1)(2n+1)}{12} + \frac{6n(n+1)}{12}\\ &= \frac{n(n+1)}{12}\left[3n(n+1) + 4(2n+1) + 6\right]\\ &= \frac{n(n+1)}{12}\left(3n^2 + 3n + 8n + 4 + 6\right)\\ &= \frac{n(n+1)}{12}\left(3n^2 + 11n + 10\right)\\ &= \frac{n(n+1)(n+2)(3n+5)}{12} \end{aligned}\]

Part (a)(ii)

We evaluate \(\displaystyle\sum_{r=5}^{n-1}(r+2)(r+3)^2\).

Replace \(r\) with \(r - 2\) (so the summand becomes \(r(r+1)^2\) and the limits shift: \(r=5\to r=7\), \(r=n-1\to r=n+1\)): \[\begin{aligned} \sum_{r=5}^{n-1}(r+2)(r+3)^2 &= \sum_{r=7}^{n+1} r(r+1)^2\\ &= \sum_{r=1}^{n+1} r(r+1)^2 - \sum_{r=1}^{6} r(r+1)^2\\ &= \frac{(n+1)(n+2)(n+3)(3n+8)}{12} - \frac{6(7)(8)(23)}{12}\\ &= \frac{(n+1)(n+2)(n+3)(3n+8)}{12} - 644 \end{aligned}\]

Many students tried replacing \(r\) with \(r+2\) in the expression from part (i), which does not give the required form. The cleaner approach is to replace \(r\) with \(r-2\) in part (ii)'s sum to reduce it to \(\sum r(r+1)^2\).

Part (b)

The recurrence is \(u_{n+1} = \dfrac{1}{1-u_n}\), with \(u_1 = 2\).

\[\begin{aligned} u_2 &= \frac{1}{1-2} = -1\\ u_3 &= \frac{1}{1-(-1)} = \frac{1}{2}\\ u_4 &= \frac{1}{1-\frac{1}{2}} = 2 \end{aligned}\]

The sequence is periodic with period 3: \(2,\,-1,\,\tfrac{1}{2},\,2,\,-1,\,\tfrac{1}{2},\,\ldots\)

Since \(2025 = 3 \times 675\), we have \(u_{2025} = u_3 = \dfrac{1}{2}\).

Answer: (a)(i) \(\displaystyle\sum_{r=1}^{n} r(r+1)^2 = \dfrac{n(n+1)(n+2)(3n+5)}{12}\). (a)(ii) \(\displaystyle\sum_{r=5}^{n-1}(r+2)(r+3)^2 = \dfrac{(n+1)(n+2)(n+3)(3n+8)}{12} - 644\). (b) \(u_{2025} = \dfrac{1}{2}\).
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