1.25: Past-paper question 4

4PM1/1/November/2025 — Question 4 · 8 marks

Given that ∑r=1n(4r+A)=n2(4n−6)\displaystyle \sum_{r=1}^n (4r + A) = \frac{n}{2}(4n - 6)
where A\displaystyle A is a constant
(a) show that A=−5\displaystyle A = -5
(3)
Hence, or otherwise,
(b) (i) evaluate ∑2160(4r−5)\displaystyle \sum_{21}^{60} (4r - 5)
(2)
(ii) find the greatest value of n\displaystyle n such that
(3)
∑r=1n(4r−5)<3418\displaystyle \sum_{r=1}^n (4r - 5) < 3418