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Devin
MaxineIf the question excludes the endpoint 160 then: I guess that you could reason as follows:
2 + 158 = 160
4 + 156 = 160
...
78 + 82 = 160
There are 39 pairs that add up to 160 plus one 80 that is not paired.
(39 * 160) + 80 = 6320
If the question is meant to include the endpoint 160 ,then (40 * 160) + 80 = 6480
To find the sum of all even numbers from 1 to 160, we first need to identify the even numbers in that range. The even numbers in this range are 2, 4, 6, ..., 158, 160. We can use the formula for the sum of an arithmetic series: sum = (n/2)(first term + last term), where n is the number of terms. In this case, n = 80 (since there are 80 even numbers from 2 to 160), the first term is 2, and the last term is 160. Plugging these values into the formula, we get: sum = (80/2)(2 + 160) = 40 * 162 = 6480. Therefore, the sum of all even numbers from 1 to 160 is 6480.
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What is the sum of all the first 100 even numbers?
The sum of all the first 100 even numbers is 10,100.
From zero to 16 is all the odd numbers the sum of even numbers?
If this question means "in the interval 0 to 16 inclusive, is the sum of the odd numbers the same as the sum of the even numbers ?" then the answer is no. The sum of the even numbers is eight more than the sum of the odd ones.
Write an algorithm to print sum of all even numbers?
Start print "the sum of all even numbers is infinite" end
Consider the numbers from 1-500 inclusive what is the difference between the sum of all of the even numbers and the sum of all the odd numbers?
501
The sum of two even numbers?
The sum of two even numbers is always an even number.
