Problem 50
Question
Find the sum. $$ \sum_{j=0}^{5} 7\left(\frac{3}{2}\right)^{j} $$
Step-by-Step Solution
Verified Answer
The sum is approximately 145.47.
1Step 1: Understand the Problem
We need to find the sum of a geometric series given by \( \sum_{j=0}^{5} 7\left(\frac{3}{2}\right)^{j} \). This involves identifying the first term and the common ratio.
2Step 2: Identify the Geometric Sequence
The series is geometric with the first term \( a = 7 \) and the common ratio \( r = \frac{3}{2} \). We have 6 terms ranging from \( j = 0 \) to \( j = 5 \).
3Step 3: Use the Formula for the Sum of a Finite Geometric Series
The sum of the first \( n+1 \) terms of a geometric series is given by: \[S_n = a \frac{r^{n+1} - 1}{r - 1}\]Here, \( n = 5 \), \( a = 7 \), and \( r = \frac{3}{2} \).
4Step 4: Calculate the Sum Substituting the Values
Substitute the given values into the formula: \[S_5 = 7 \cdot \frac{\left(\frac{3}{2}\right)^{6} - 1}{\frac{3}{2} - 1}\]Calculate \( \left(\frac{3}{2}\right)^{6} \) and solve for \( S_5 \).
5Step 5: Compute the Values
First compute \( \left(\frac{3}{2}\right)^{6} = \frac{729}{64} \). Substitute back, \[S_5 = 7 \cdot \frac{\frac{729}{64} - 1}{\frac{3}{2} - 1}\]
6Step 6: Final Calculation
Calculate the remaining expressions.\[S_5 = 7 \cdot \frac{\frac{729 - 64}{64}}{\frac{1}{2}} = 7 \cdot \frac{665}{64} \cdot 2 \]Further solve to find \( S_5 \).
7Step 7: Simplify and Obtain Final Answer
Complete the arithmetic:\[S_5 = 7 \cdot \frac{665}{32} = 7 \times 20.78125 = 145.46875\]Therefore, the sum of the series is approximately 145.47.
Key Concepts
finite geometric seriescommon ratiosum formulageometric sequence
finite geometric series
A finite geometric series is a series that has a specific number of terms. Unlike infinite series, where terms continue indefinitely, finite series come to an end. In the given exercise, the finite geometric series is expressed as \( \sum_{j=0}^{5} 7\left(\frac{3}{2}\right)^{j} \). Here, the series starts with \( j = 0 \) and ends with \( j = 5 \), which means it includes six terms in total. Each term is derived from the geometric sequence formula, with the entire series reflecting a pattern where every term has a consistent multiplying factor, known as the common ratio.
common ratio
The common ratio in a geometric sequence or series is the factor by which we multiply each term to get the subsequent term. It is an essential characteristic that defines the nature of a geometric series. In the exercise, the series \( 7\left(\frac{3}{2}\right)^{j} \) has a common ratio \( r = \frac{3}{2} \). This means each term is 1.5 times larger than the preceding term. Understanding the common ratio helps us predict the nature and values of subsequent terms in the sequence, and it plays a crucial role when using the sum formula for calculating the total of the series.
sum formula
To find the sum of a finite geometric series, we use the sum formula, which allows us to compute the total value of all terms in the series without adding each one individually. The sum formula is given by:
- \[ S_n = a \frac{r^{n+1} - 1}{r - 1} \]
geometric sequence
A geometric sequence is a list of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sequence given in the exercise starts with the first term, \( a = 7 \), and follows a consistent pattern due to the common ratio \( r = \frac{3}{2} \). The terms of this sequence are:
- \( 7 \)
- \( 7 \times \frac{3}{2} \)
- \( 7 \times \left(\frac{3}{2}\right)^2 \)
- ... up to \( 7 \times \left(\frac{3}{2}\right)^{5} \)
Other exercises in this chapter
Problem 49
Simplify using the Binomial Theorem. Show that \((1.01)^{100}>2 .\) [Hint: Note that \((1.01)^{100}=(1+0.01)^{100},\) and use the Binomial Theorem to show that
View solution Problem 50
\(49-54\) . A partial sum of an arithmetic sequence is given. Find the sum. $$ -3+\left(-\frac{3}{2}\right)+0+\frac{3}{2}+3+\cdots+30 $$
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\(49-54\) . Use a graphing calculator to evaluate the sum. $$ \sum_{k=1}^{100}(3 k+4) $$
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Simplify using the Binomial Theorem. Show that \(\left(\begin{array}{l}{n} \\ {0}\end{array}\right)=1\) and \(\left(\begin{array}{l}{n} \\ {n}\end{array}\right)
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