Summation of Series: \(\sum r^3\) result; index shift — JPJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
(a) The first three terms of a sequence are given by \(u_{1}=9\), \(u_{2}=27\) and \(u_{3}=55\). Given that \(u_{n}\) is a quadratic polynomial in \(n\), find \(u_{n}\) in terms of \(n\).
(b) It is given that \(\displaystyle \sum_{r=1}^{n} r^{3} = \frac{n^{2}(n+1)^{2}}{4}\) and \(u_{r}=2r^{3}+5\).
- Find \(\displaystyle \sum_{r=1}^{n} u_{r}\).
- Hence, or otherwise, find \(\displaystyle \sum_{r=2}^{n} \bigl(2(r+2)^{3}+5\bigr)\).
- Explain why the series \(\displaystyle \sum_{r=1}^{\infty} u_{r}\) does not converge.
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(a) [3 marks]
Let \(u_{n} = an^{2}+bn+c\). Then \[\begin{aligned} u_{1} &= a+b+c = 9 \quad\text{--- (1)}\\ u_{2} &= 4a+2b+c = 27 \quad\text{--- (2)}\\ u_{3} &= 9a+3b+c = 55 \quad\text{--- (3)} \end{aligned}\]
Using GC: \(a = 5\), \(b = 3\), \(c = 1\). \[ u_{n} = 5n^{2}+3n+1. \]
(b)(i) [2 marks] \[\begin{aligned} \sum_{r=1}^{n}\!\left(2r^{3}+5\right) &= 2\sum_{r=1}^{n}r^{3}+\sum_{r=1}^{n}5\\ &= 2\!\left(\frac{n^{2}(n+1)^{2}}{4}\right)+5n\\ &= \frac{n^{2}(n+1)^{2}}{2}+5n \end{aligned}\]
(b)(ii) [3 marks]
Method 1: Substitute and use part (i) \[\begin{aligned} \sum_{r=2}^{n}\!\left(2(r+2)^{3}+5\right) &= \left(2(2+2)^{3}+5\right)+\cdots+\left(2(n+2)^{3}+5\right)\\ &= \sum_{r=4}^{n+2}\!\left(2r^{3}+5\right)\\ &= \sum_{r=1}^{n+2}\!\left(2r^{3}+5\right)-\sum_{r=1}^{3}\!\left(2r^{3}+5\right)\\ &= \frac{(n+2)^{2}(n+3)^{2}}{2}+5(n+2)-\left[\frac{3^{2}(4)^{2}}{2}+5(3)\right]\\ &= \frac{(n+2)^{2}(n+3)^{2}}{2}+5n-77 \end{aligned}\]
Method 2: Split sums of cubes and constants \[\begin{aligned} \sum_{r=2}^{n}\!\left(2(r+2)^{3}+5\right) &= \sum_{r=4}^{n+2}\!\left(2r^{3}+5\right)\\ &= 2\!\left[\sum_{r=1}^{n+2}r^{3}-\sum_{r=1}^{3}r^{3}\right]+\sum_{r=4}^{n+2}5\\ &= 2\!\left[\frac{(n+2)^{2}(n+3)^{2}}{4}-\frac{3^{2}(4)^{2}}{4}\right]+5(n-1)\\ &= \frac{(n+2)^{2}(n+3)^{2}}{2}+5n-77 \end{aligned}\]
(b)(iii) [1 mark]
From (i), \(\displaystyle\sum_{r=1}^{n}u_{r} = \frac{n^{2}(n+1)^{2}}{2}+5n\).
As \(n \to \infty\), \(\dfrac{n^{2}(n+1)^{2}}{2} \to \infty\) and \(5n \to \infty\).
Hence, \(\displaystyle\sum_{r=1}^{n}u_{r} = \frac{n^{2}(n+1)^{2}}{2}+5n \to \infty\). The series \(\displaystyle\sum_{r=1}^{\infty}u_{r}\) does not converge.