The population of a local species of dragonfly can be found using an infinite geometric series where \(a_1 = 42\) and the common ratio is \(\frac{3}{4}\).

Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

A. \(\sum_{i=1}^{\infty} \left(\frac{3}{4}\right) 42^{i-1}\); the sum is divergent.
B. \(\sum_{i=1}^{\infty} \left(\frac{3}{4}\right) 42^{i-1}\); the sum is 168.
C. \(\sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}\); the sum is divergent.
D. [tex]\(\sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}\)[/tex]; the sum is 168.



Answer :

To solve this problem, we begin by understanding the components of the infinite geometric series given in the question. We have:

- \(a_1 = 42\) as the first term.
- \(r = \frac{3}{4}\) as the common ratio.

We need to represent the sum of the series in sigma notation and also determine if the series converges or diverges, then calculate the sum if it converges.

### Step-by-Step Solution

1. Representing in Sigma Notation:

In an infinite geometric series, each term after the first can be found by multiplying the previous term by the common ratio \(r\). Therefore, the series can be written as:

[tex]\[ a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots \][/tex]

For our given values:
[tex]\[ 42, 42 \left(\frac{3}{4}\right), 42 \left(\frac{3}{4}\right)^2, 42 \left(\frac{3}{4}\right)^3, \ldots \][/tex]

This series can be expressed in sigma notation as:
[tex]\[ \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1} \][/tex]

2. Determining Convergence:

An infinite geometric series converges if the absolute value of the common ratio is less than 1, i.e., \(|r| < 1\). Here, our common ratio \(r = \frac{3}{4}\), which satisfies this condition since \(|\frac{3}{4}| < 1\).

3. Calculating the Sum:

The sum \(S\) of an infinite convergent geometric series can be calculated using the formula:
[tex]\[ S = \frac{a_1}{1 - r} \][/tex]

Substituting the given values:
[tex]\[ S = \frac{42}{1 - \frac{3}{4}} = \frac{42}{\frac{1}{4}} = 42 \times 4 = 168 \][/tex]

### Final Answer:

The sigma notation for the series is:
[tex]\[ \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1} \][/tex]

The sum is:
[tex]\[ 168 \][/tex]

Therefore, the correct option is:
[tex]\[ \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}; \text{ the sum is 168 }. \][/tex]

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