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

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

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.