Problem 53
Question
Find the sum of each infinite geometric series. $$ \sum_{i=1}^{\infty} 7(0.4)^{i-1} $$
Step-by-Step Solution
Verified Answer
The sum of the infinite geometric series is \(\frac{35}{3}\).
1Step 1: Identify the first term (a)
The first term of the series can be found by substituting the first value of the series. In this case, replace \(i\) with 1 in the given series: \(a = 7(0.4)^{1-1} = 7(0.4)^0 = 7(1) = 7\)
2Step 2: Identify the common ratio (r)
The common ratio is the factor multiplied to get from one term to the next in the series. Here, the common ratio \(r\) is 0.4, as it is given in the series.
3Step 3: Use the formula for the sum of an infinite geometric series
The formula for the sum of an infinite geometric series is \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.
4Step 4: Substitute the known values into the formula
Using the formula \(S = \frac{a}{1-r}\), substitute \(a = 7\) and \(r = 0.4\): \(S = \frac{7}{1-0.4} = \frac{7}{0.6}\)
5Step 5: Simplify the expression
Now simplify the expression: \(S = \frac{7}{0.6} = \frac{7 \times 10}{0.6 \times 10} = \frac{70}{6} = \frac{35}{3}\).
Key Concepts
geometric seriesseries sum formulacommon ratiofirst term calculationinfinite series solution
geometric series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant value, called the common ratio. For example, in the series 7, 2.8, 1.12, ... each term is obtained by multiplying the previous term by 0.4. A geometric series can be finite or infinite. If it's infinite, it continues forever. Geometric series have many applications in fields like finance, physics, and computer science.
series sum formula
The formula to find the sum of an infinite geometric series is very useful. It's given by
\( S = \frac{a}{1-r} \)
where:
\( S = \frac{a}{1-r} \)
where:
- \textbf{a}: the first term of the series
- \textbf{r}: the common ratio
common ratio
The common ratio (\textbf{r}) in a geometric series is the constant value that each term is multiplied by to get the next term. It's calculated as:
\( r = \frac{a_2}{a_1} \)
where \( a_2 \) is the second term and \( a_1 \) is the first term. In our example,
\( a_2 = 2.8 \) and \( a_1 = 7 \),
thus \( r = \frac{2.8}{7} = 0.4 \). The common ratio tells us how quickly the terms are getting smaller or larger.
\( r = \frac{a_2}{a_1} \)
where \( a_2 \) is the second term and \( a_1 \) is the first term. In our example,
\( a_2 = 2.8 \) and \( a_1 = 7 \),
thus \( r = \frac{2.8}{7} = 0.4 \). The common ratio tells us how quickly the terms are getting smaller or larger.
first term calculation
Identifying the first term (\textbf{a}) in a geometric series sets the foundation for calculating the sum of the series. To find the first term, substitute the first index value into the series expression. For our given series:
\( \text{The series is} \ \ \sum_{i=1}^{otegthinspaceotegthiotegthinspaceotegthinspaceotegthinspace\frac{otegthinspaceoteg}7}{ otegthinspaceototegthinspace (0.4)^{i-1} \;&n \}-1} \ \ (0.4)^{i-1} = ...of| \ = ...of| \ $ \ a_{s} = 7(1} = (0.4)^0 = ...1 = \ \ ing the first term = \The series' term \ a \ = \ \). :::::::::::::}}::::}} Prplayers ing the indice = }\0. \ \term.Originally found the finds:
\( \text{The series is} \ \ \sum_{i=1}^{otegthinspaceotegthiotegthinspaceotegthinspaceotegthinspace\frac{otegthinspaceoteg}7}{ otegthinspaceototegthinspace (0.4)^{i-1} \;&n \}-1} \ \ (0.4)^{i-1} = ...of| \ = ...of| \ $ \ a_{s} = 7(1} = (0.4)^0 = ...1 = \ \ ing the first term = \The series' term \ a \ = \ \). :::::::::::::}}::::}} Prplayers ing the indice = }\0. \ \term.Originally found the finds:
infinite series solution
Solving an infinite geometric series involves substituting the known values for the first term (\textbf{a}) and the common ratio (\textbf{r}) into the sum formula:
\( S = \frac{a}{1-r} \)
For our specific series:
\( S = \frac{7}{1 - 0.4} = \frac{7}{0.6} \)
Simplify the expression to find:
\( S = \frac{7}{0.6} = \frac{7 \times 10}{0.6 \times 10} = \frac{70}{6} = \frac{35}{3} \)
Therefore, the sum of the infinite series is \( \frac{35}{3} \).
\( S = \frac{a}{1-r} \)
For our specific series:
- \textbf{a} = 7
- \textbf{r} = 0.4
\( S = \frac{7}{1 - 0.4} = \frac{7}{0.6} \)
Simplify the expression to find:
\( S = \frac{7}{0.6} = \frac{7 \times 10}{0.6 \times 10} = \frac{70}{6} = \frac{35}{3} \)
Therefore, the sum of the infinite series is \( \frac{35}{3} \).
Other exercises in this chapter
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