Answer:
Option 3 - \(S=\sum_{i=1}^{\infty} 36(\frac{1}{2})^i-1\) ; the sum is 72.
Step-by-step explanation:
Given : The population of a local species of dragonfly can be found using an infinite geometric series where \(a_1 = 36\) and the common ratio is one half.
To find : Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.
Solution :
Geometric series is \(a+ar+ar^2+ar^3+.......\)
The formula of sum of geometric infinite series is
\(S=\sum_{k=0}^{\infty} ar^k=\frac{a}{1-r}\)
In the given geometric series,
\(a_1 = 36\) and \(r=\frac{1}{2}\)
According to question,
\(S=\sum_{i=1}^{\infty} 36(\frac{1}{2})^i-1=\frac{36}{1-\frac{1}{2}}\)
\(S=\sum_{i=1}^{\infty} 36(\frac{1}{2})^i-1=\frac{36}{\frac{1}{2}}\)
\(S=\sum_{i=1}^{\infty} 36(\frac{1}{2})^i-1=72\)
Therefore, Option 3 is correct.
\(S=\sum_{i=1}^{\infty} 36(\frac{1}{2})^i-1\) ; the sum is 72.
