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Chapter 8: Problem 56

Express each sum using summation notation. Use a lower limit of summation of your choice and \(k\) for the index of summation. $$6+8+10+12+\dots+32$$

Short Answer

Expert verified
The summation notation for the series 6+8+10+12+...+32 is \(\sum_{k=1}^{14} (4 + 2k)\).

Step by step solution

01

Identify the common difference

In any arithmetic sequence, you first want to identify the common difference between each term. Here, each term increases by 2. So, the common difference (\(d\)) is 2.
02

Find the general form of the term

The general form of an arithmetic sequence is written as \(a_r = a_1 + (k - 1) \cdot d\), where \(a_r\) is the r-th term we want to determine, \(a_1\) is the first term, \(k\) is the index of summation, and \(d\) is the common difference. Here \(a_1 = 6\) (the first term of the series) and \(d = 2\) (the common difference). Therefore, the general form of the term would be \(a_r = 6 + (k - 1) \cdot 2 = 4 + 2k\).
03

Set the limits for summation

The exercise states that we can use a lower limit of summation of our choice. The series starts at 6 (which corresponds to \(k = 1\)), and ends at 32 (which corresponds to \(k = 14\) as 32 corresponds to the 14th term in the series). Therefore, the limits for the summation will be from 1 to 14 as there are 14 terms in this series.
04

Write the summation notation

The final step is to write the summation notation using the formula found in Step 2 and the limits from Step 3. This becomes: \(\sum_{k=1}^{14} (4 + 2k)\).

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