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.

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caylus caylus · Answer 1 · 2016-05-22T13:04:15-04:00

Hello,

\(\lim_{n \to \infty} \sum_{i=0}^{i=n} a_1*(\frac{3}{4})^i =42*4=168\)

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wagonbelleville wagonbelleville · Answer 2 · 2019-03-18T02:04:43-04:00

Answer:

The sum in sigma form is \(\sum_{i=1}^{\infty}42(\frac{3}{4})^{i}\)

The upper limit of the population is 168.

Step-by-step explanation:

We are given that,

Population of dragonfly is represented by the series with,

First term, \(a_{1}=42\)

Common ratio, \(r=\frac{3}{4}\)

So, we see that,

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

That is, \(\sum_{i=1}^{\infty}42(\frac{3}{4})^{i}\)

Now, the infinite sum of the series is \(S=\frac{a_1}{1-r}\)

So, the sum is \(S=\frac{42}{1-\frac{3}{4}}\)

i.e. \(S=\frac{42\times 4}{4-3}\)

i.e. \(S=\frac{168}{1}\)

Thus, the upper limit of the population is 168.