Question

Suppose a population of 50 crickets doubles in size every 4 months. How many crickets will there be after 5 years?

Answer 1

1,638,350

Explanation:

This is a Geometric Progression problem. We can use the formula`a_n = a_1 xx q^(n-1)` to find any particular value in the series. https://www.math10.com/en/algebra/geometric-progression.html#calc

We need to adjust the end value (5 years) to the increment value (4 months). The number of increments is `(12 "months"/"year" )*(5"years")/(4"months")) = 15` incremental periods.

`a_15 = 50 xx 2^(15-1) = 50 xx 2^14 = 819200` on the last increment.

The formula for the sum* of the first n numbers of a geometric series is:
`S_n = (a_1 – a_nq)/(1 -  q) = a_1 xx (1 – q^n)/( 1 - q)`

Geometric Progression: 50, 100, 200, 400, 800, 1600, 3200, 6400, 12800, 25600, 51200, 102400, 204800, 409600, 819200

The 15th term: 819200

Sum of the first 15 terms: 1,638,350

Answer 2

`1,638,400` crickets

Explanation:

This is an example of exponential growth.

We can use the formula `f(t)=a*(1+r)^t`, where `a` is the initial amount, `r` is the growth rate (represented by a decimal), and `t` is the time interval.

Let's say `t` represents the number of months. Convert `5` years to months.

`5 "years" * (12 "months")/(1 "year") = 60 "months"`

Since the crickets are doubling, they are basically increasing by `100%`. When written as a decimal, `100%=1`, so `r=1`.

So, `a=50` (the initial number of crickets), `r=1`, and `t=60`. Plugging these values in, we get

`f(60)=50*(1+1)^(60)`

However, it is important to note that the crickets are doubling every four months, not every month. So we must divide `t` by `4`.

`f(60)=50*(1+1)^(60/4)`

Now, we can solve for `f(60)`.

`f(60)=50*2^(15)`

`f(60)=50*32768`

`f(60)=1,638,400`

So, at the end of `5` years, there will be `1,638,400` crickets.