Problem 11 - Algebra for College Students

Question

Find the sum of each series. $$\sum_{i=1}^{5} 2^{-i}$$

Step-by-Step Solution

Answer

\(\frac{31}{32}\)

Step 1: Identify the series

The series in question is \(\sum_{i=1}^{5} 2^{-i}\). This represents the sum of the terms when the exponent runs from 1 to 5, with each term being \(2^{-i}\).

Step 2: Write out the first few terms

List the first few terms of the series: \(2^{-1}, 2^{-2}, 2^{-3}, 2^{-4}, 2^{-5}\). This translates to \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}\).

Step 3: Sum the terms manually

Add each of the terms together step by step: \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}\). To simplify this, it helps to use a common denominator: 32. Convert each fraction: \(\frac{16}{32} + \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{1}{32}\). Adding these together: \(16 + 8 + 4 + 2 + 1 = 31\), thus the sum is \(\frac{31}{32}\).

Key Concepts

geometric series

A geometric series is one where each term is obtained by multiplying the previous term by a constant factor known as the common ratio. In our problem, the series is \(\sum_{i=1}^{5} 2^{-i}\). Each term here decreases by a factor of \(2\).
For instance:

• 1st term: \(2^{-1} = \frac{1}{2}\)
• 2nd term: \(2^{-2} = \frac{1}{4}\)
• 3rd term: \(2^{-3} = \frac{1}{8}\)
• 4th term: \(2^{-4} = \frac{1}{16}\)
• 5th term: \(2^{-5} = \frac{1}{32}\)

Understanding geometric series helps us to comprehend how a sequence of numbers progresses and the nature of their sums.

fraction addition

Adding fractions involves finding a common denominator. This allows us to sum the numerators directly while keeping the denominator the same.
For instance, to add \(\frac{1}{2}\) and \(\frac{1}{4}\), convert each fraction to have a common denominator of 4:
\(\frac{1}{2} = \frac{2}{4}\) Next, add the numerators: \(\frac{2}{4} + \frac{1}{4} = \frac{3}{4}\)
In our problem, summing all terms required converting each fraction to a denominator of 32:

• \(\frac{1}{2} = \frac{16}{32}\)
• \(\frac{1}{4} = \frac{8}{32}\)
• \(\frac{1}{8} = \frac{4}{32}\)
• \(\frac{1}{16} = \frac{2}{32}\)
• \(\frac{1}{32} = \frac{1}{32}\)

Adding these fractions gives us \(\frac{31}{32}\) as the sum.

partial sums

A partial sum is simply the sum of the first 'n' terms of a series. In our case, it's the sum of the first 5 terms:

• \(\frac{1}{2}\)
• \(\frac{1}{4}\)
• \(\frac{1}{8}\)
• \(\frac{1}{16}\)
• \(\frac{1}{32}\)

The total is computed by adding each fraction together, ensuring they are converted to a common denominator for ease:

• \(\frac{16}{32} + \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{1}{32} = \frac{31}{32}\)

The concept of partial sums is useful as it helps in understanding the behavior of a series as more terms are added.