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<blockquote data-quote="Thanee" data-source="post: 177883" data-attributes="member: 478">Back in school (about the 7th year in school, don't remember exactly) I derived the equation 1+2+3+...+n=n(n+1)/2 without knowing about it yet. It really helps to understand why it works like this.There are two possible cases, if you add up number 1...n, case one being, that n is even, and case two being, that n is odd.Case One (even n):Group those n numbers in pairs - 1 and n, 2 and (n-1), 3 and (n-2), and so on. All those pairs add up to (n+1) obviously, since you always add one and subtract one in the sum. There are (also obvious) (n/2) of such pairs, one half the number of single numbers in the sum.Therefore we got n/2 pairs of value (n+1), or (n/2)*(n+1) in total, thus1+2+3+...+n=(n/2)*(n+1)=(1/2)*n*(n+1)=n(n+1)/2 (for even n).Case Two (odd n):With a little trick you can use the same derivate of Case One. Just add 0 to the sum of numbers, making it an even number of numbers again. 1+2+3+...+n=0+1+2+3+...+n.Now we can again group those (n+1) numbers in pairs - 0 and n, 1 and (n-1), 2 and (n-2), and so on. All those pairs add up to n and there are ((n+1)/2) of such pairs, therefore the result is slightly different.We got ((n+1)/2) pairs of value n, or ((n+1)/2)*n in total, thus0+1+2+3+...+n=((n+1)/2)*n=(1/2)*(n+1)*n=n(n+1)/2 (for odd n).Since the set of natural numbers only consists of even or odd numbers, this proves the equation for all natural numbers.Of course, nowadays I'm working on much more complex mathematical problems, but I still think this is kinda fascinating, how easy this highly useful equation works out, if you put a little thought into it.ByeThanee</blockquote>

[QUOTE="Thanee, post: 177883, member: 478"] Back in school (about the 7th year in school, don't remember exactly) I derived the equation 1+2+3+...+n=n(n+1)/2 without knowing about it yet. It really helps to understand why it works like this. There are two possible cases, if you add up number 1...n, case one being, that n is even, and case two being, that n is odd. Case One (even n): Group those n numbers in pairs - 1 and n, 2 and (n-1), 3 and (n-2), and so on. All those pairs add up to (n+1) obviously, since you always add one and subtract one in the sum. There are (also obvious) (n/2) of such pairs, one half the number of single numbers in the sum. Therefore we got n/2 pairs of value (n+1), or (n/2)*(n+1) in total, thus 1+2+3+...+n=(n/2)*(n+1)=(1/2)*n*(n+1)=n(n+1)/2 (for even n). Case Two (odd n): With a little trick you can use the same derivate of Case One. Just add 0 to the sum of numbers, making it an even number of numbers again. 1+2+3+...+n=0+1+2+3+...+n. Now we can again group those (n+1) numbers in pairs - 0 and n, 1 and (n-1), 2 and (n-2), and so on. All those pairs add up to n and there are ((n+1)/2) of such pairs, therefore the result is slightly different. We got ((n+1)/2) pairs of value n, or ((n+1)/2)*n in total, thus 0+1+2+3+...+n=((n+1)/2)*n=(1/2)*(n+1)*n=n(n+1)/2 (for odd n). Since the set of natural numbers only consists of even or odd numbers, this proves the equation for all natural numbers. Of course, nowadays I'm working on much more complex mathematical problems, but I still think this is kinda fascinating, how easy this highly useful equation works out, if you put a little thought into it. Bye Thanee [/QUOTE]

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