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Chapter 11: Problem 42

Find the sum of the first 80 positive even integers.

Short Answer

Expert verified
The sum of the first 80 positive even integers is 6480.

Step by step solution

01

Identify the first term

The first positive even integer is \( 2 \), so the first term, \( a_1 \), is \( 2 \).
02

Identify the common difference

The difference between any two consecutive even numbers is \( 2 \), so the common difference \( d \) is \( 2 \). However, in this case, we don't need the common difference directly for our calculation, but it is useful for understanding the structure of the sequence.
03

Identify the 80th term

Since we know the first term \( a_1 \) and common difference \( d \), we can find the 80th term using the formula \( a_n = a_1 + (n-1)*d \). Here, \( n = 80 \), \( a_1 = 2 \), \( d = 2 \). Substituting these into the formula gives: \( a_{80} = 2 + (80-1)*2 =160 \). Thus, the 80th term, \( a_{80} \), is 160.
04

Find the sum

Now we have all we need to apply the formula for the sum of an arithmetic series: \( S_n = \frac{n}{2} (a_1 + a_n) \). Substituting the values: \( S_{80} = \frac{80}{2} (2 + 160) = 40 * 162 = 6480 \). Thus, the sum of the first 80 positive even integers is 6480.

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