How to get SUM for series S = 1 + 2 + 3 + .. + (n-1) + n?

Question

Equals S = (n * ( n + 1 )) / 2

I want to understand how the above formula is arrived at.

Another series

K = 1 + 2 + 4 + .. .. .. + 2 (n-2) + 2 (n-1) ----- (Equation 2.a)

Can be solved as follows

Multiply Equation 2.a with 2 on both sides

2K = 2 + 4 + 8 + .. .. .. + 2 (n-1) + 2 (n) ----- (Equation 2.b)

now subtract equation 2.a from 2.b and we get the following

K = 2 (n) - 1

I want a similar solution (i.e steps) for the Equation 1.

On internet I found good answers at

http://www.9math.com/book/sum-first-n-natural-numbers

http://en.wikipedia.org/wiki/Mathematical_induction

but I want a solution without induction.

( note in Equation 2.a and 2.b , read 2 (n-2) as 2 power (n-2) and

2 (n-1) as 2 power (n-1)

Answer 1

First you write

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

The second you write:

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

Add between them and obtain

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

You see that (n+1) issued n times

Then

2S=n(n+1)

Then

S=n(n+1)/2

Answer 2

Future reference: If you want to do powers/exponents/indices (whatever you may call them), use the "^" symbol.  So 2 to the power of (n-2) would be written 2^(n-2).

I don't know any proof that works like the one for the geometric series... however, here is a similar trick:

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

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

Add them to get:

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

The number of terms on the right-hand side didn't change, so there are still n terms there.

2S = n * (n+1)

S = (n * (n+1)) / 2.