Answer to Question #136792 in Discrete Mathematics for Promise Omiponle

Question

Suppose that the number of bacteria in a colony triples every hour. Let Bn denote the number of bacteria in the colony after n hours.
(a) Set up a recurrence relation for Bn.
(b) If 100 bacteria are used to begin a new colony, how many bacteria will be in the colony in 10 hours?

Answer 1

Solution to (a)

Let the starting value at hour 0 be B₀

B₁ =3B₀

B₂=3B₁

B₃ =3B₂

B_n=3B_n-1

Thus the recurrence relation for B_{n }is

B_n=3B_n-1

Solution to (b)

B₀= 100

$B_n = 3B_{n-1}$

$$\therefore B_{10} = 3B_9 = 3(3B_8)=3(9B_7)=3(27B_6)$$ $$=3(81B_5)=3(243B_4)=3(729B_3)=3(2187B_2)$$ $$=3(6561B_1)=3(19683B_0)$$ $$\therefore B _{10} =59049B_0$$

$$=59049(100)=5904900$$

Answer: 5904900 bacteria