Possible Duplicate:
Proof for formula for sum of sequence $1+2+3+\ldots+n$?
I was just messing around in Haskell:
let fnc x = (sum [1..x], x * (x - 1) / 2)
fnc 12
>> (78.0, 66.0)
(I was just messing around after reading The Mythical Man Month: "Intercommunication is worse. If each part of the task must be separately coordinated with each other part/ the effort increases as n(n-I)/2.")
Then I noticed this:
[fnc x | x <- [1..20]]
>> [(1.0,0.0),(3.0,1.0),(6.0,3.0),(10.0,6.0),(15.0,10.0),(21.0,15.0),(28.0,21.0),(36.0,28.0),(45.0,36.0),(55.0,45.0),(66.0,55.0),(78.0,66.0),(91.0,78.0),(105.0,91.0),(120.0,105.0),(136.0,120.0),(153.0,136.0),(171.0,153.0),(190.0,171.0),(210.0,190.0)]
(Scratching head and thinking 'Hey, y is always the x of the last step')
And so I wrote:
let fnt x = (sum [1..x], (x + 1) * x / 2)
fnt 3
>> (6.0,6.0)
My math skills are horrible but I wanted to transform one side to the other. I failed. miserably.
Why are those equations equal? Are there any interesting benefits mathematically or computationally in defining sum as (x + 1) * x / 2 ?
