Chapter 8.2, Problem 50E

To determine

To find: how many sterile flies are there after n days and what happens in the long run.

Answer to Problem 50E

  $S_n=\frac{N}{1-s}.$

Explanation of Solution

Given information: To control an agriculture pest called the medfly, N sterilized male

flies are released into the general fly population every day. If s is the proportion of these sterilized flies that survive a given day, then $N{s}^k$ will survive for k days

Calculation:

N sterilized male flies are released into the general fly population every day.

  $S_1=N$ .

s is the proportion of these sterilized flies that survive a given day, then the number of

flies that survive any given day is:

  $N{s}^k$ .

So, the number of sterilized male flies that there are in two day is ,

  $\begin{array}{l}{S}_2=N+Ns\\ S_3=N+Ns+N{s}^2\\ \mathrm{..........}\\ \mathrm{...........}\\ S_n=N+Ns+N{s}^2+\mathrm{......}N{s}^{n-1}\\ S_n={\displaystyle \sum _{n=1}^{\infty }N{s}^{n-1}}\end{array}$

So, this is a geometric series with ratio r=s and a=N.

So, sum of this geometric series is ,

  $\begin{array}{l}{S}_n={\displaystyle \sum _{n=1}^{\infty }N{s}^{n-1}}=\frac{N}{1-s}.\\ S_n=\frac{N}{1-s}.\end{array}$