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TMUA 2017 · Paper 2 · Question 15 of 20

TMUA 2017 Paper 2 Question 15

Sequences and series — Iterated rules · finding the cycle. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 2Sequences and seriesIterated rules · finding the cycle6 options
The function f is defined on the positive integers as follows: f(1) = 5, and for n ≥ 1, f(n+1) = 3f(n) + 1   if  f(n)  is odd f(n+1) = 12f(n)   if  f(n)  is even The function g is defined on the positive integers as follows: g(1) = 3, and for n ≥ 1, g(n+1) = g(n) + 5   if  g(n)  is odd g(n+1) = 12g(n)   if  g(n)  is even What is the value of f(1000) − g(1000)?
  1. A−6
  2. B−5
  3. C1
  4. D2
  5. E4
  6. F8
Show the answer and worked solution
answer · D
  1. A−6
  2. B−5
  3. C1
  4. D2
  5. E4
  6. F8
Generate terms until each sequence repeats. For f: 5,  16,  8,  4,  2,  1,  4,  2,  1,  …, so from n = 4 onwards it cycles through 4,  2,  1 with period 3, taking the value 4 whenever n leaves remainder 1 on division by 3. As 1000 = 3×333 + 1, f(1000) = 4. For g: 3,  8,  4,  2,  1,  6,  3,  …, a cycle of length 6 starting at n = 1; as 1000 = 6×166 + 4, g(1000) = g(4) = 2. The difference is 4 − 2 = 2.