Further Pure Mathematics 4PM1 / Sequences and Series / 1.161.16: Show that (b) Hence, or otherwise, evaluate Given that (c) find the value of4PM1/1/June/2024 — Question 3 · 8 marksMark as done · Save for later(a) Show that ∑r=1n(5r−3)=n2(5n−1)\displaystyle \sum_{r=1}^n (5r - 3) = \frac{n}{2}(5n - 1)r=1∑n(5r−3)=2n(5n−1)(3)(b) Hence, or otherwise, evaluate ∑r=3160(5r−3)\displaystyle \sum_{r=31}^{60} (5r - 3)r=31∑60(5r−3)(2)Given that ∑r=1n(5r−3)=3783\displaystyle \sum_{r=1}^n (5r - 3) = 3783r=1∑n(5r−3)=3783(c) find the value of n\displaystyle nn(3)▸ Mark scheme← 1.24: 4PM1/2R/June/2025 — Question 41.17: 4PM1/1/June/2024 — Question 8 →
