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Find the sum of the numbers 1 to 100?

I have finals tomorrow and I'm trying to do some pre-sleep studying. I knew how to do this in the beginning of the year but I havent used it in a while. Help!!

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  • 1 decade ago
    Favorite Answer

    It's a sneaky formula:

    n*(n+1)/2

    n here would be 100.

    so

    100*101/2 = 5050

    You can always remember this by testing it with a simple one. Remember that 1+2+3 = 6. If you can remember that 3*4/2 = 6, you'll be all set. (I know I always have to remind myself if it is (n+1) or (n-1) in there.)

    Good luck tomorrow!

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  • 4 years ago

    1

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  • 1 decade ago

    There is a well known story about Karl Friedrich Gauss when he was in

    elementary school. His teacher got mad at the class and told them to

    add the numbers 1 to 100 and give him the answer by the end of the

    class. About 30 seconds later Gauss gave him the answer.

    The other kids were adding the numbers like this:

    1 + 2 + 3 + . . . . + 99 + 100 = ?

    But Gauss rearranged the numbers to add them like this:

    (1 + 100) + (2 + 99) + (3 + 98) + . . . . + (50 + 51) = ?

    If you notice every pair of numbers adds up to 101. There are 50

    pairs of numbers, so the answer is 50*101 = 5050. Of course Gauss

    came up with the answer about 20 times faster than the other kids.

    In general to find the sum of all the numbers from 1 to N:

    1 + 2 + 3 + 4 + . . . . + N = (1 + N)*(N/2)

    That is "1 plus N quantity times N divided by 2."

    Hope this helps.

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  • 3 years ago

    2

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  • Anonymous
    1 decade ago

    5050.

    You can acheive this result by adding up the numbers like this.

    1+100, 2+99, 3+98........ all the way to...... 50+51. So you see there are 50 pairs of 101. So 50 times 101 equals 5050.

    Hope I helped!

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  • Anonymous
    1 decade ago

    haha i saw this question so many times

    make couples of 1 and 99, 2 and 98, 3 and 97, etc until you reach 50, so that there will be 49 100's, then add 50 and 100, so you get 5050

    not sure, i did this long time ago

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  • Anonymous
    1 decade ago

    Let's say we have numbers 1 through n.

    To find the sum of those numbers, it would be n * (n+1) / 2

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