Summation of Series: GP: \(\sum r^2\) standard result — DHS 2025 H2 Math Prelim Paper 1
What this question tests
Question
A series is given by \(\displaystyle\sum_{r=1}^{n} 2(4-3x)^r\) where \(x\) is constant.
- Explain why this is a geometric series. Determine the range of values of \(x\) for the sum to infinity of this series to exist.
- Using \(x = 1\), and given that \(\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n}{6}(n+1)(2n+1)\), find \(\displaystyle\sum_{r=0}^{n-1}\bigl[2(4-3x)^r + (r+1)(2r+5)\bigr]\), leaving your answer in the form \(n(an^2 + bn + c)\), where \(a\), \(b\) and \(c\) are constants to be determined.
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(a) The general term is \(u_r = 2(4-3x)^r\). The ratio of consecutive terms is: \[ \frac{u_{r+1}}{u_r} = \frac{2(4-3x)^{r+1}}{2(4-3x)^r} = 4-3x \quad \text{(constant)} \] Since the common ratio \(4-3x\) is constant, the series is a geometric series.
For the sum to infinity to exist, \(|4-3x| < 1\): \[\begin{aligned} -1 < 4-3x &< 1 \\ -5 < -3x &< -3 \\ 1 < x &< \frac{5}{3} \end{aligned}\]
Range: \(1 < x < \dfrac{5}{3}\)
(b) With \(x = 1\), the geometric sum gives \(2(4-3)^r = 2(1)^r = 2\). So: \[\begin{aligned} \sum_{r=0}^{n-1}\bigl[2(4-3x)^r + (r+1)(2r+5)\bigr] &= \sum_{r=0}^{n-1}2 + \sum_{r=0}^{n-1}(r+1)(2r+5) \\ &= 2n + \sum_{r=0}^{n-1}(r+1)(2r+5) \end{aligned}\]
Substituting \(r \to r-1\) (i.e. replace \(r\) with \(r-1\) so limit goes \(r=1\) to \(n\)): \[\begin{aligned} \sum_{r=0}^{n-1}(r+1)(2r+5) &= \sum_{r=1}^{n}r(2r+3) = \sum_{r=1}^{n}(2r^2 + 3r) \\ &=2 \sum_{r=1}^{n}r^2 + 3 \sum_{r=1}^{n}r\\ &= 2\cdot\frac{n}{6}(n+1)(2n+1) + 3\cdot\frac{n(n+1)}{2} \\ &= \frac{n(n+1)(2n+1)}{3} + \frac{3n(n+1)}{2} \end{aligned}\]
Therefore the full sum is: \[\begin{aligned} &= 2n + \frac{n(n+1)(2n+1)}{3} + \frac{3n(n+1)}{2} \\ &= n\!\left(2 + \frac{(n+1)(2n+1)}{3} + \frac{3(n+1)}{2}\right) \\ &= n\!\left(\frac{2}{3}n^2 + \frac{5}{2}n + \frac{23}{6}\right) \end{aligned}\]