Suppose a newly formed cell takes two minutes to reach maturity and from that point on can reproduce every minute. Starting with one cell when n = 0, let a(n) be the number of cells after n minutes.

Question

a. Write out the first five terms of the sequence {a(n)} infinity, n=0

b. Write a formula for an in terms of a(n-1) and a(n-2)

c. Let b(n) = a(n+1)/a(n). Write a recursive formula for b(n) in terms of b(n-1).

d. Find limit as n approaches infinity - b(n)

Answer 1

This is a Fibonacci sequence.

a) 1,1,2,3,5

b) a(n) = a(n-1)+a(n-2), n>2

c) b(n) = a(n+1)/a(n)

b(n-1) = a(n)/a(n-1)

a(n-1) = a(n)/b(n-1)

b(n) = (a(n)+a(n-1))/a(n) = 1 + (a(n)/b(n-1))/a(n) = 1 +1 /b(n-1)

d) lim n→∞ 1+1/b(n-1)

b(2) = 1+1/1 = 2

b(3) = 1+1/2 = 3/2

b(4) = 1+2/3 = 5/3

b(5) = 1+3/5 = 8/5

b(6) = 1+5/8 = 13/8

b(7) = 1+8/13 = 21/13

b(8) = 1+13/21 = 34/21

Looks like it's approaching 1.61803398874989 = (1+√5)/2 = the golden ratio

b(n) = 1 +1/b(n-1)

b(n)b(n-1)-b(n)-1 = 0

When n→∞, n and n-1 are essentially the same.

b²-b-1=0

Use quadratic equation

b = (1+√(1+4))/2 = (1+√5)/2 --> the golden ratio