Problem 42 - Algebra and Trigonometry

Question

Find the sum of the first 80 positive even integers.

Step-by-Step Solution

Answer

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

Step 1: Identify the first term

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

Step 2: 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.

Step 3: 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.

Step 4: 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.