Question

Use a graphing utility to find the sum of each geometric sequence. $$ \sum_{n=1}^{15} 4 \cdot 3^{n-1} $$

Short answer

28697812

Step-by-step solution

Identify the first term and common ratio

The first term of the geometric sequence, denoted as \(a\), is the coefficient of the given formula when \(n = 1\). For the formula given, \(a = 4 \cdot 3^{0} = 4\). The common ratio, denoted as \(r\), is the base of the exponential part, which is 3 in this case.

Write the sum formula for a geometric series

The formula for the sum of the first \(n\) terms of a geometric sequence is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \(S_n\) is the sum of the sequence, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Substitute the known values into the sum formula

In this case, \(a = 4\), \(r = 3\), and \(n = 15\). Substituting these values into the formula gives: \[ S_{15} = 4 \frac{1 - 3^{15}}{1 - 3} \]

Simplify the formula

Simplify the denominator first: \(1 - 3 = -2\). Now compute the numerator: \(1 - 3^{15}\). Substituting these into the formula gives: \[ S_{15} = 4 \frac{1 - 14348907}{-2} \]

Compute the final result

Calculate \(1 - 14348907 = -14348906\). Now, \[ S_{15} = 4 \frac{-14348906}{-2} = 4 \cdot 7174453 = 28697812 \]

Key concepts

Geometric Series

A geometric series is a sum of the terms in a geometric sequence. Each term after the first is obtained by multiplying the preceding term by a fixed, non-zero number called the common ratio. In the given problem, a geometric series is expressed as: \[\begin{equation} \textstyle \frac{1}{4} \[\[\begin{align*} \textstyle 4 + 4 \times 3^{1} + 4 \times 3^{2} + 4 \times 3^{3} + \textstyle ... + 4 \times 3^{14} \ otag\end{align*}\]\] \end{equation}\] Notice that each term is obtained by multiplying the previous term by 3, making 3 the common ratio. Understanding this pattern is essential for working with geometric series.

Sum Formula

The sum formula for a geometric series allows you to find the total sum of the first n terms without needing to add each term individually. This formula is written as: \[\begin{equation} S_n = a \frac{1-r^n}{1-r} \end{equation}\] Here, \textbf{S\textsubscript{n}} is the sum of the first n terms, \textbf{a} is the first term, and \textbf{r} is the common ratio. Let's break it down further:

• a is the first term of the sequence.
• r is the common ratio between the terms.
• n is the number of terms to be added in the sequence.

In this problem, using the formula makes the calculation easier, as opposed to manually adding all terms.

Common Ratio

The common ratio in a geometric sequence is the factor by which we multiply each term to get the next term. In the given problem \[\begin{equation} {sum_n (\theta)} = 4 \times 3^{n-1} \end{equation}\] , the common ratio is 3. This can be identified as follows: after the first term (which is4), each subsequent term is the previous term multiplied by 3:

• First term: 4
• Second term: 4 \times 3 = 12
• Third term: 12 \times 3 = 36
• and so on...

Recognizing and correctly identifying the common ratio is crucial for solving problems involving geometric sequences and series.