In a pest eradication program, "N" sterilized male flies are released into a general population each day.

It is estimated that 90% of these flies will survive a give data.

A) Show that the number of sterilzed flies in population "n" days after the program begun is:

N+(0.9)N+N(0.9)^2+N(0.9)^3+......+N(0.9)^(n-1),
(geometric sequence)

B) If long range goal of the program is to keep 20,000 sterilized male flies in the population, how many flies should be released each day?

A) just check the population after each day

1: 90% survive; a new batch is added.
Total is .90N+N
2: 90% of 1st day's group survive: .90*.90
90% of 2nd batch survive; N more added
total: .90*.90N + .90N + N
So, after n days, you can see the proposed sequence.

B) You want the limit to be 20000, so

N/(1-.90) = 20000
N =200,000

A) To prove that the number of sterilized flies in the population "n" days after the program begins is given by the geometric sequence:

N + (0.9)N + N(0.9)^2 + N(0.9)^3 + ... + N(0.9)^(n-1)

We can use the formula for the sum of a geometric sequence:

S = a * (1 - r^n) / (1 - r)

where:
- S is the sum of the sequence
- a is the first term of the sequence
- r is the common ratio of the sequence
- n is the number of terms in the sequence

In this case, the first term (a) is N, and the common ratio (r) is 0.9. We want to find the sum after "n" terms, so:

S = N * (1 - (0.9)^n) / (1 - 0.9)

We can simplify it further:

S = N * (1 - 0.9^n) / 0.1

Therefore, the number of sterilized flies in the population "n" days after the program begins is:

N + (0.9)N + N(0.9)^2 + N(0.9)^3 + ... + N(0.9)^(n-1) = N * (1 - 0.9^n) / 0.1

B) If the long-range goal of the program is to keep 20,000 sterilized male flies in the population, we need to set the sum S equal to 20,000 and solve for N:

N * (1 - 0.9^n) / 0.1 = 20000

Multiplying both sides by 0.1:

N * (1 - 0.9^n) = 2000

Dividing both sides by (1 - 0.9^n):

N = 2000 / (1 - 0.9^n)

This equation gives you the number of flies that should be released each day (N) based on the number of days (n) and the desired population size.

A) To prove that the number of sterilized flies in the population "n" days after the program began is a geometric sequence, we need to show that each term is obtained by multiplying the previous term by a constant ratio.

Let's start by analyzing the pattern in the number of sterilized flies in the population each day:

- Day 0: N sterilized flies
- Day 1: 0.9N sterilized flies (90% survive)
- Day 2: (0.9)(0.9)N = (0.9)^2N sterilized flies
- Day 3: (0.9)(0.9)(0.9)N = (0.9)^3N sterilized flies
- ...
- Day n-1: (0.9)^(n-1)N sterilized flies

From this pattern, we can see that each term is obtained by multiplying the previous term by a constant factor of 0.9. Therefore, the number of sterilized flies in the population "n" days after the program began can be expressed as a geometric sequence:

N + (0.9)N + (0.9)^2N + (0.9)^3N + ... + (0.9)^(n-1)N

B) To determine the daily release of flies to maintain 20,000 sterilized male flies in the population in the long run, we can set up an equation using the geometric sequence formula:

N + (0.9)N + (0.9)^2N + (0.9)^3N + ... = 20,000

This is an infinite geometric series, but we can calculate the sum of the series by using the formula for an infinite geometric series:

Sum = a / (1 - r)

where:
- a is the first term, which is N
- r is the common ratio, which is 0.9

Substituting these values into the formula, we get:

N / (1 - 0.9) = 20,000

Simplifying the equation, we have:

N / 0.1 = 20,000

Multiply both sides by 0.1:

N = 20,000 * 0.1

N = 2,000

Therefore, to maintain 20,000 sterilized male flies in the population in the long run, you should release 2,000 flies each day.