Question

write the sigma notation for the geometric series from the following finite geometric sequence: 3, 9, 27, 81, 243. (1 point) \\(\sum_{n=1}^{5} 5(3)^{n - 1}\\) \\(\sum_{n=1}^{3} 5(3)^{n - 1}\\) \\(\sum_{n=1}^{5} 3(3)^{n - 1}\\) \\(\sum_{n=1}^{3} 3(3)^{n - 1}\\)

Explanation:

Step1: Identify the first term and common ratio

The geometric sequence is \(3, 9, 27, 81, 243\). The first term \(a_1 = 3\), and the common ratio \(r=\frac{9}{3}=3\). The number of terms \(n = 5\) (since there are 5 terms: 3, 9, 27, 81, 243).

Step2: Recall the formula for the sum of a finite geometric series in sigma notation

The formula for the \(n\)-th term of a geometric sequence is \(a_n=a_1r^{n - 1}\), and the sum in sigma notation is \(\sum_{k = 1}^{n}a_1r^{k - 1}\). Here, \(a_1 = 3\), \(r = 3\), and \(n=5\). So the sigma notation should be \(\sum_{n = 1}^{5}3(3)^{n - 1}\).

Answer:

\(\sum_{n = 1}^{5}3(3)^{n - 1}\) (the third option: \(\boldsymbol{\sum_{n=1}^{5} 3(3)^{n - 1}}\))