What is the sum of the first 100 positive even integers?

Answer 1

use this formula: S=(N/2)(F+L) S= the sum of the first 100 positive even integers N= the number of terms F= first term L=last term Since we are using the first 100 positive integers we will replace "N" with 100. so it will look like this : S=(100/2)(F+L) Next we must substitute the "F" with the first positive term which is the number 2. So now it looks like this: S=(100/2)(2+L) We use the number 2 because our positive integers are 2,4,6,8,10,12...... you get the picture. Now we must substitute the "L" with the last positive term which is 200. As you probably guessed it will look like this: S=(100/2)(2+200) We use the number 200 because it is the last positive even integer. If we used 202 then that would have meant we used the 101th positive even integer and we don't want to do that. == == We should solve the numbers inside the parenthesis first. S=(100/2)(2+200) 100 divided by 2 = 50 So: S=50(2+200) 2+200=202 That means: S=50(202) 50 x 202 = 10,100 Finally: S=10,100 There you go! Wasn't that easy? Note: This formula only works if you are looking for the sum of the first 100 positive or negative, even or odd integers. It sucks cause this formula is the Bomb right? Oh well