<p>A series is given by \(\sum_{r=1}^{n} 2(4-3x)^r\) where \(x\) is constant.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Explain why this is a geometric series. Determine the range of values of \(x\) for the sum to infinity of this series to exist.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Using \(x=1\), and given that \(\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=0}^{n-1} [2(4-3x)^r + (r+1)(2r+5)]\), leaving your answer in the form of \(n(an^2+bn+c)\), where \(a\), \(b\) and \(c\) are constants to be determined.</span><span class="inline-marks">[4]</span></div></li></ol>

A series is given by \(\sum_{r=1}^{n} 2(4-3x)^r\) 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 \(\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=0}^{n-1} [2(4-3x)^r + (r+1)(2r+5)]\), leaving your answer in the form of \(n(an^2+bn+c)\), where \(a\), \(b\) and \(c\) are constants to be determined.

2025 DHS P1 Q2

2025 DHS P1 Q2