Problem 50
Question
Use a graphing utility to find the sum of each geometric sequence. $$ \sum_{n=1}^{15} 4 \cdot 3^{n-1} $$
Step-by-Step Solution
Verified Answer
28697812
1Step 1: 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.
2Step 2: 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.
3Step 3: 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: \[ S_{15} = 4 \frac{1 - 3^{15}}{1 - 3} \]
4Step 4: Simplify the formula
Simplify the denominator first: \(1 - 3 = -2\). Now compute the numerator: \(1 - 3^{15}\). Substituting these into the formula gives: \[ S_{15} = 4 \frac{1 - 14348907}{-2} \]
5Step 5: Compute the final result
Calculate \(1 - 14348907 = -14348906\). Now, \[ S_{15} = 4 \frac{-14348906}{-2} = 4 \cdot 7174453 = 28697812 \]
Key Concepts
Geometric SeriesSum FormulaCommon Ratio
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.
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...
Other exercises in this chapter
Problem 49
Find the value of \(\left(\begin{array}{l}5 \\\ 0\end{array}\right)\left(\frac{1}{4}\right)^{5}+\left(\begin{array}{l}5 \\\ 1\end{array}\right)\left(\frac{1}{4}
View solution Problem 49
Expand each sum. \(\sum_{k=1}^{n}(k+2)\)
View solution Problem 50
The entries in the Pascal Triangle can, for \(n \geq 2\), be used to determine the number of \(k\) -sided figures that can be formed using a set of \(n\) points
View solution Problem 50
Expand each sum. \(\sum_{k=1}^{n}(2 k+1)\)
View solution