the following two lines indicate another way to derive the formula for the sum of the first n integers by rearranging the terms in the sum. Fill in the details.

Question

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

Answer 1

...
= (1+n)+(1+n)+(1+n)+... for a total of n/2 such binomials
= (n/2)(n+1
or
n(n+1)/2 or (1/2)(n)(n+1) or (n^2 + n)/2