AQUARIUS

  Joyrex said:

  mcbpete said:

  azatoth said:

btw, is there a simple formula for adding up consecutive numbers?

(n * (n+1)) / 2

 

eg 1+2+3+4+5 = (5*6)/2 = 15

 

 

That's neat... I'll have to teach my kids that one, unless they already know it!

There's a common story that attributes this technique to the ridiculous ingenious Carl Gauss, from when he was in primary school. The teacher was looking for a bit of a break so he told the class to add up all the numbers from 1 to 100, figuring this would take them a good while. Gauss shocked teacher by coming up with the right answer within a minute. He realized that each pair of numbers, taken from each end, forms the same total: 1 + 100 = 2 + 99 = 3 + 98 ... there are 50 pairs, so it's just (1 + 100) * (100/2) = 101 * 50 = 5050.

I made my own alternate equation on the sum of any number to any other number. so far i havent met anyone who has seen the equation before. It would be awesome if i can get my name added to it. So I dunno if I wanna post it on the internet just yet. It's much more complicated than Carl's as his only goes from 1 to any other number.

Yeah man same with me and my music I just don't wanna release it yet, even though its soooooo much more complicated than RDJ/Autechre/everyone!!

Just giving you a hard time, PM me it and I'll be your best friend! Pinkie swear!!!

if there is a list of n consecutive numbers to add up, between a and b, where a<b, then the sum is [(b+1)*(b/2)]-[a*((a-1)/2)] in all cases.

so, for the sum of the consecutive integers between 6 and 9, for example, it's [(9+1)*(9/2)]-[6*(5/2)] = 45-15 = 30

is that the proof that you "discovered"?