TTMUA Lab
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. A6
  2. B5
  3. C1
  4. D2
  5. E4
  6. F8
Show the answer and worked solution
answer · D
  1. A6
  2. B5
  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.