Question

What is the sum of the even numbers from 1 to 500, inclusive?

Answer 1

The sum of all odd numbers I trough 99 is 9,801

Answer 2

Thus, the required sum of the even numbers from 1 to 500 is 62750.

Given,
To determine the sum of the even numbers from 1 to 500, inclusive

What is arithmetic progression?

Arithmetic progression is the series of numbers that have common differences between adjacent values is same. An arithmetic progression is represented by the expression is $a_n = a +(n-1)d$, wherer a = first term, d = common difference and n = nth term.

Since from 1 to 500 there are 500/2 = 250 even numbers, where the first term ( a ) is 2 and the common difference ( d ) is 2.
The sum of n = 250 even numbers is given as,

Sn = n/2 [2a + (n-1)d]

Where n is the number of elements, a is the first term and d is a common difference.

$S_{250}= 250/2[2*2+(250-1)2]$
      = 125 [ 4 + 498 ]
      = 62750

Thus, the required sum of the even numbers from 1 to 500 is 62750.

Learn more about arithmetic progression here: https://brainly.com/question/20334860
#SPJ2