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Q: The sum of the first 100 positive even numbers?

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The sum of the first 100 positive even numbers is 10,100.

10,100

-998 = if its negative number 100 = if its positive numbers

You do not need a conjecture; you can calculate the answer. The answer is 10,100

The sum of the first 100 even numbers is 10,100

All numbers are used in all known languages, just spelt differntly, so no-one knows,sorry

The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)

The sum of all the first 100 even numbers is 10,100.

2 + 4 + 6 + 8 + 10 = 30 In general, the sum of the first n even numbers is n(n+1). Replacing n with 5 we have 5x6 which equals 30 as we found by direct calcuation. The formula is nice when you have many numbers. For example, finding the sum of the first 100 positive even numbers would be a PAIN! But using the formula is it simply 100x101.

10100.

There are 25 of them.

2,550

We have to assume that you're talking about whole numbers. The sum is 5,050 .

50%

n*(n+1)=sum 100*(100+1)=10100

The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.

even numbers 1-100

The sum of the 25 primes less than 100 is even (it is 1060).The first prime number is 2. Every other prime number is odd.So, of the 25 primes less than 100, one is even and 24 are odd.Any two odd numbers sum to an even number. Therefore, an even number of odd numbers has an even sum. So the sum of 24 odd numbers is even.Therefore, the sum of the 25 primes less than 100 is even.

The trick is the # of terms squared. (the term - 100 - multiplied by itself - 1002 = 100 x 100)

Positive 3-digit numbers are any numbers between 100 and 999. They are not negative (no "-" sign in front) and are three digits.

Should be 50! Every odd integer is 1 less than the corresponding even integer and there are 50 of each in 100...

There are 16 positive numbers smaller than 100 that are directly divisible by 6.

Two solutions immediately spring to mind:Run over all the numbers in the range and if the number is even print it;run over the numbers starting with the first even number in the range using a step of 2 (so that only even numbers are considered) and print the numbers.

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

The sum of the first 100 numbers, excluding zero, is 5,001.

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