Question

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

Answer 1

Hello,


`\lim_(n \to \infty) \sum_(i=0)^(i=n) a_1*((3)/(4))^i =42*4=168`

Answer 2

The sum in sigma form is
`\sum_(i=1)^(\infty)42((3)/(4))^(i)`

The upper limit of the population is 168.

Explanation:

We are given that,

Population of dragonfly is represented by the series with,

First term,
`a_(1)=42`

Common ratio,
`r=(3)/(4)`

So, we see that,

The sum in sigma form is given by
`\sum_(i=1)^(\infty)a_(1)r^(i)`

That is,
`\sum_(i=1)^(\infty)42((3)/(4))^(i)`

Now, the infinite sum of the series is
`S=(a_1)/(1-r)`

So, the sum is
`S=(42)/(1-(3)/(4))`

i.e.
`S=(42* 4)/(4-3)`

i.e.
`S=(168)/(1)`

Thus, the upper limit of the population is 168.