1+2+3+...+n=(1+n)+(2+(n-1))+(3+(n-2))+...

=(1+n)+(1+n)+(1+n)+...

How many copies of n+1 are in there in the second line?
with n=6 and n=7

1 answer

Since the n numbers are divided up in pairs, I'd say n/2 copies.

Hence the formula for the sum: n(n+1)/2

Technically, that's [n+1]/2 where [] is the greatest-integer function.