Question

Suppose that the number of bacteria in a colony triples every hour.

a) Set up a recurrence relation for the number of bacteria after n hours have elapsed.

b) If 100 bacteria are used to begin a new colony, how many bacteria will be in the colony in 10 hours?

Short answer

(a) The recurrence relation for number of bacteria after n hours is ${\mathrm{X}}_{\mathrm{n}}=3{\mathrm{X}}_{\mathrm{n}-1}$.

(b) The number of bacteria in new colony is 5,904,900 .

Step-by-step solution

Determination of recurrence relation for the number of bacteria(a)Given Data:

${\mathrm{X}}_{\mathrm{n}}=3{\mathrm{X}}_{\mathrm{n}-1}$The number of hours is n = 10 hr

The initial number of bacteria in new colony is $X_0=100$

The recurrence relation is found by observing the steps for some fixed number of times and recurrence relation formed on the basis of repeation of pattern.

Suppose the initial number of bacteria at time n = 0 are $X_0$. The number of bacteria for first hour is given as:

$X_1=3X_0$

The number of bacteria for second hour is given as:

$X_2=3X_1$

The number of bacteria for third hour is given as:

$X_3=3X_2$

The number of bacteria for ${\left(\mathbf{n}\mathbf{-}\mathbf{1}\right)}^{\mathrm{th}}$hour is given as:

$X_{n-1}=\left(3\right)X_{n-2}$

Now on the above observations number of bacteria for ${\mathrm{n}}^{\mathrm{th}}$hour is given as:

$X_n=3X_{n-1}$

Therefore, the recurrence relation for number of bacteria after n hours is

Determination of number of bacteria(b)

The explicit formula for above recurrence is given as:

$X_n=(3{)}^n{X}_0$

Substitute all the values in the above equation.

$\begin{array}{r}{X}_{10}=\left(3{)}^{10}\right(100)\\ X_{10}=5,904,900\end{array}$

Therefore, the number of bacteria in new colony is 5,904,900 .