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TMUA 2017 · Paper 1 · Question 17 of 20

TMUA 2017 Paper 1 Question 17

Sequences and series — Integral-defined sequence · sum of an arithmetic series. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Sequences and seriesIntegral-defined sequence · sum of an arithmetic series5 options
The two functions F(n) and G(n) are defined as follows for positive integers n: F(n) = 1n∫0n (n − x) dx G(n) = ∑r=1n F(r) What is the smallest positive integer n such that G(n) > 150?
  1. A22
  2. B23
  3. C24
  4. D25
  5. E26
Show the answer and worked solution
answer · D
  1. A22
  2. B23
  3. C24
  4. D25
  5. E26
Evaluate F first, treating n as a constant inside the integral: ∫0n (n−x) dx = [nx − x22]0n = n22, so F(n) = n2. Then G(n) = 12∑r=1n r = n(n+1)4. The condition G(n) > 150 becomes n(n+1) > 600. At n = 24 the product is exactly 600, which fails the strict inequality, so the answer is n = 25.