<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">For this question, you may use these results:</div><div class="list-item-paragraph" style="text-align: center">\[\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} \quad \text{and} \quad \sum_{r=1}^{n}r^3 = \frac{n^2 (n+1)^2}{4}.\]</div><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">Show that&nbsp;\(\sum_{r=1}^{n}r(r+1)^2 = \frac{n(n+1)(n+2)(3n+5)}{12}\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Hence find \(\sum_{r=5}^{n-1}(r+2)(r+3)^2\) in terms of \(n\).<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The sequence \(u_1, u_2, u_3, \dots\) is defined by \(u_1 = 2\), \(u_{n+1} = \frac{1}{1-u_n}\), \(n \ge 1\).</div><div class="list-item-paragraph">Find the values of \(u_2\), \(u_3\) and \(u_4\). Hence find the value of \(u_{2025}\).<span class="inline-marks">[3]</span></div></li></ol>

For this question, you may use these results:

\[\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} \quad \text{and} \quad \sum_{r=1}^{n}r^3 = \frac{n^2 (n+1)^2}{4}.\]

The sequence \(u_1, u_2, u_3, \dots\) is defined by \(u_1 = 2\), \(u_{n+1} = \frac{1}{1-u_n}\), \(n \ge 1\).

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

Recurrence Relations and Summation

International Baccalaureate Higher Level Mathematics

2025 ACJC P2 Q3

2025 ACJC P2 Q3