The number of insects in a population at the start of year [tex]\( n \)[/tex] is [tex]\( P_n \)[/tex].

The number of insects in the population at the start of year [tex]\( (n+1) \)[/tex] is [tex]\( P_{n+1} \)[/tex], where
[tex]\[ P_{n+1} = k P_n \][/tex]

Given that [tex]\( k \)[/tex] has a constant value of 1.13,

(a) Find out how many years it takes for the number of insects in the population to double. You must show how you get your answer.



Answer :

To find how many years it takes for the population of insects to double when growing at a constant rate, we need to iteratively apply the given growth factor until the population reaches twice its initial size.

Let's break this down step-by-step:

1. Given Values:
- The growth factor [tex]\( k \)[/tex] is 1.13.
- The initial population [tex]\( P_n \)[/tex] is 1 (for simplicity, we'll start with 1 unit).
- The target population is 2, which is double the initial population.

2. Initialize Variables:
- Start with the initial population, [tex]\( P_n = 1 \)[/tex].
- We need a counter for the number of years, starting at 0.

3. Growth Process:
- Each year, the population grows by multiplying the current population by the growth factor [tex]\( k = 1.13 \)[/tex].
- We keep track of how many years this process takes until the population reaches or exceeds the target.

Let's describe this process step-by-step:

1. Year 0:
- Population: [tex]\( P_0 = 1 \)[/tex]
- Target: 2

2. Year 1:
- New Population: [tex]\( P_1 = P_0 \times 1.13 = 1 \times 1.13 = 1.13 \)[/tex]
- Continue to next year as [tex]\( 1.13 < 2 \)[/tex]

3. Year 2:
- New Population: [tex]\( P_2 = P_1 \times 1.13 = 1.13 \times 1.13 \approx 1.2769 \)[/tex]
- Continue to next year as [tex]\( 1.2769 < 2 \)[/tex]

4. Year 3:
- New Population: [tex]\( P_3 = P_2 \times 1.13 = 1.2769 \times 1.13 \approx 1.4439 \)[/tex]
- Continue to next year as [tex]\( 1.4439 < 2 \)[/tex]

5. Year 4:
- New Population: [tex]\( P_4 = P_3 \times 1.13 = 1.4439 \times 1.13 \approx 1.6316 \)[/tex]
- Continue to next year as [tex]\( 1.6316 < 2 \)[/tex]

6. Year 5:
- New Population: [tex]\( P_5 = P_4 \times 1.13 = 1.6316 \times 1.13 \approx 1.8437 \)[/tex]
- Continue to next year as [tex]\( 1.8437 < 2 \)[/tex]

7. Year 6:
- New Population: [tex]\( P_6 = P_5 \times 1.13 = 1.8437 \times 1.13 \approx 2.081952 \)[/tex]
- Stop here, since [tex]\( 2.081952 \)[/tex] exceeds the target of 2.

It takes 6 years for the population to double from 1 to a value exceeding 2, precisely reaching approximately 2.081952.

Therefore, the population of insects will take 6 years to double.