Chapter 13: Problem 16
Evaluate each sum. $$\sum_{k=1}^{6}\left(\frac{2}{3}\right)^{k+1}$$
Short Answer
Expert verified
\(\frac{2660}{2187}\)
Step by step solution
01
Understand the Sum
The sum we need to compute is \( \sum_{k=1}^{6} \left(\frac{2}{3}\right)^{k+1} \). This means we need to find the sum of \( \left(\frac{2}{3}\right)^{2} + \left(\frac{2}{3}\right)^{3} + \left(\frac{2}{3}\right)^{4} + \left(\frac{2}{3}\right)^{5} + \left(\frac{2}{3}\right)^{6} + \left(\frac{2}{3}\right)^{7} \).
02
Evaluate First Few Terms
Evaluate the first few terms manually: \((\frac{2}{3})^{2} = \frac{4}{9}\), \((\frac{2}{3})^{3} = \frac{8}{27}\), and \((\frac{2}{3})^{4} = \frac{16}{81}\).
03
Recognize the Geometric Series
Notice that the series is geometric with the first term \(a = (\frac{2}{3})^2\) and the common ratio \(r = \frac{2}{3}\).
04
Use the Formula for Sums of Finite Geometric Series
The formula for the sum of the first \(n\) terms of a geometric series is \(S_n = a \frac{1-r^n}{1-r}\) where \(a\) is the first term and \(r\) is the common ratio. Here, \(a = (\frac{2}{3})^2\) and \(r = \frac{2}{3}\), and we want the sum of the first 6 terms, which means \(n = 6\).
05
Substitute Values into the Formula
Substitute \(a = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\), \(r = \frac{2}{3}\), and \(n = 6\) into the formula to get:\[ S_6 = \frac{4}{9} \frac{1-(\frac{2}{3})^6}{1-\frac{2}{3}}. \] First calculate \((\frac{2}{3})^6 = \frac{64}{729} \).
06
Simplify the Expression
Plug into the formula: \[ S_6 = \frac{4}{9} \frac{1-\frac{64}{729}}{\frac{1}{3}} = \frac{4}{9} \times \frac{665}{729} \times 3. \]Simplify the expression to find \(S_6\). Simplify each step: \(\frac{665}{729} \times 3 = \frac{1995}{729}\). Then compute \(\frac{4}{9} \times \frac{1995}{729}\), which simplifies further to get the answer.
07
Calculate Final Answer
Multiply to obtain the final result: \(\frac{4}{9} \times \frac{1995}{729}\).This gives the finalized value of the sum.Ultimately, the computed sum equals approximately \(\frac{2660}{2187}\), which is accurate when checked.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum of Geometric Series
To find the sum of a geometric series, there is a handy formula. A geometric series is a series with a constant ratio between successive terms. This specific formula makes it easy to sum all terms up to a certain number. The formula for the sum of the first \( n \) terms of a geometric series is:
- \( S_n = a \frac{1-r^n}{1-r} \)
- \( S_n \): the sum of the series
- \( a \): the first term in the series
- \( r \): the common ratio between the terms
- \( n \): the number of terms
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 number called the common ratio. In the sequence, each term grows by the factor of this ratio.
- Example: A sequence like \( 2, 6, 18, 54 \) has a common ratio of 3.
- In our exercise, the sequence starts with \((\frac{2}{3})^2\) and has a common ratio of \( \frac{2}{3} \).
Series Evaluation
Evaluating a series is all about calculating the sum of its numbers. For a geometric series, this process becomes much straightforward using the geometric series formula.To evaluate a series, here’s what you usually do:
- Identify the first term \( a \).
- Find the common ratio \( r \).
- Decide how many terms \( n \) you want to sum up.
- Plug these values into the formula \( S_n = a \frac{1-r^n}{1-r} \) to get the result.