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?

Solution to (a)

Let the starting value at hour 0 be B0

B1 =3B0

B2=3B1

B3 =3B2

Bn=3Bn-1

Thus the recurrence relation for Bn is

Bn=3Bn-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