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Magazine Chapter 14: Problem 70
Write out and evaluate each sum. $$ \sum_{k=3}^{7} \frac{k}{2^{k}} $$
Short Answer
Expert verified
The sum is \(\r\frac{119}{128} \).
Step by step solution
01
Understanding the Summation
The given summation is \(\r \sum_{k=3}^{7} \frac{k}{2^{k}} \). This means you need to find the sum of the expression \(\r\frac{k}{2^{k}}\) as k varies from 3 to 7.
02
Write Out Each Term
Identify the terms by substituting k with the integers from 3 to 7: \( \frac{3}{2^{3}} + \frac{4}{2^{4}} + \frac{5}{2^{5}} + \frac{6}{2^{6}} + \frac{7}{2^{7}} \).
03
Evaluate Each Term
Calculate each individual term: \(\r \frac{3}{2^{3}} = \frac{3}{8} \, \frac{4}{2^{4}} = \frac{4}{16} = \frac{1}{4} \), \(\r \frac{5}{2^{5}} = \frac{5}{32} \), \(\r \frac{6}{2^{6}} = \frac{6}{64} = \frac{3}{32} \), and \(\r \frac{7}{2^{7}} = \frac{7}{128} \).
04
Find a Common Denominator
Convert these fractions to have a common denominator of 128 for easier addition:\r \(\r \frac{3}{8} = \frac{48}{128} \), \frac{1}{4} = \frac{32}{128} \, \frac{5}{32} = \frac{20}{128} \, \frac{3}{32} = \frac{12}{128} \, and \(\r\frac{7}{128} \).
05
Add the Terms
Add all the converted fractions: \(\r\frac{48}{128} + \frac{32}{128} + \frac{20}{128} + \frac{12}{128} + \frac{7}{128} = \frac{119}{128} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Summation Notation
Summation notation is a concise way of representing the addition of a sequence of numbers. The Greek letter Sigma (鈭�) is used to denote summation. For example, \(\r\sum_{k=3}^{7} \frac{k}{2^{k}}\) tells us to sum the expression \(\r\frac{k}{2^{k}}\) as k ranges from 3 to 7.\r\rThis notation helps us simplify complex additions and visualize patterns in series. Each component in the summation has a specific function:\r
- \r
- The lower limit (3) tells us where to start summing. \r
- The upper limit (7) indicates where to stop summing. \r
- The expression \(\r\frac{k}{2^{k}}\) is what we sum at each step. \r
Evaluating Series
Evaluating a series means finding the total sum of all terms within the defined range of the summation notation. Here's how we approached it for our given series step by step:\r\rFirst, identify each term by substituting k values from 3 to 7 into the expression \(\r\frac{k}{2^{k}}\).\r
- \r
- For k=3: \(\r\frac{3}{2^{3}} = \frac{3}{8}\) \r
- For k=4: \(\r\frac{4}{2^{4}} = \frac{1}{4}\) \r
- For k=5: \(\r\frac{5}{2^{5}} = \frac{5}{32}\) \r
- For k=6: \(\r\frac{6}{2^{6}} = \frac{3}{32}\) \r
- For k=7: \(\r\frac{7}{2^{7}} = \frac{7}{128}\) \r
- \r
- \(\r\frac{3}{8} = \frac{48}{128}\) \r
- \(\r\frac{1}{4} = \frac{32}{128}\) \r
- \(\r\frac{5}{32} = \frac{20}{128}\) \r
- \(\r\frac{3}{32} = \frac{12}{128}\) \r
- \(\r\frac{7}{128}\) \r
Fractions
Fraction manipulation is key when working with summations involving rational numbers. Here, understanding fractions helped us evaluate our series step by step.\r\rFractions consist of a numerator (top number) and a denominator (bottom number). When adding fractions, they need a common denominator. Here's why:\r
- \r
- Common Denominator: It allows fractions to be combined easily. For \(\r\frac{3}{8}\) and \(\r\frac{1}{4}\), converting to \(\r\frac{48}{128}\) and \(\r\frac{32}{128}\) makes addition straightforward. \r
- Multiplying Fractions: For conversion, sometimes the numerator and denominator are both multiplied by the same number, ensuring the fraction's value doesn鈥檛 change. \r
- Simplifying Fractions: Finally, always reduce fractions to the simplest form if possible for clarity. \r
