Further Pure Mathematics 4PM1 / Sequences and Series / 1.111.11: Past-paper question 14PM1/1/June/2023 — Question 1 · 5 marksMark as done · Save for later(a) Show that ∑r=1n(3r+2)=n2(3n+7)\quad (a) \text{ Show that } \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7)(a) Show that r=1∑n(3r+2)=2n(3n+7)(3)(b) Hence, or otherwise, evaluate ∑r=1040(3r+2)(b) \text{ Hence, or otherwise, evaluate } \sum_{r=10}^{40} (3r + 2)(b) Hence, or otherwise, evaluate r=10∑40(3r+2)(2)▸ Mark scheme← 1.19: 4PM1/2R/June/2024 — Question 81.14: 4PM1/1/November/2023 — Question 7 →
