Math Summation Proof....Simple

1 + 2 + 3 + .... + i - 2 + i - 1 = [( i - 1)i]/2

Prove with gory detail please.

Let

K = 1 + 2 + 3 + .... + (i - 2) + (i - 1)
now, use the commutative property
of addition to write

K = (i-1)+(i-2)+(i-3)+...+2+1

Note that K+K =2K

and line by line

1+(i-1) =i
2+(i-2) =i
.
.
(i-1) +1 =i

NOTE
this is i-1 different i's

so

K+K = 2K = i times i-1


If

2K = i(i-1)

divide both sides by 2

...................i(i-1)
K = sum = ▬▬▬▬
......................2

the series is

1,2,3,4.....i-1

now..1+i-1=i
and 2+i-2=i

similarly select the rth term and the rth term from the last
that is r+(i-r)=i
so the sum of each such pair is i
there are 2 cases now
1. if i-1 is even

then there are (i-1)/2 numbers of such pairs
since the sum of each pair is i
the total sum is s=i*(i-1)/2

case 2 if i-1 is odd

there (i-2)/2 no of such pairs

and the middle number in the series will be without a pair { consider 1,2,3,4,5,6,7} for example
the middle number is given by {(i-1)+1}/2=i/2

so the sum is sum of pairs + i/2

s=(i-2)*i/2+i/2=(i^2-2i+i)/2=(i^2-i)/2
=i*(i-1)/2

thus in both cases sum=i*(i-1)/2