Factorisation help - 5* for best answer

How is
(1+2+3+...+n)/n

factorised into
[n(n+1)/2]/n ??

Thanks

P.S. Steps would be appreciated as I would really like to get my head around this, not just get to the answer...

This isn't factorization, it's a formula.

There are a couple of different ways to find this formula:

The usual way (intuitive way) is to note that if you add the first and last number, then the second and second to last number, then third, fourth, etc. you get the same sum:

1 + 2 + 3 + 4 + 5 + 6
--> 1 + 6 = 7, 2 + 5 = 7, and 3 + 4 = 7

So just add the first and last number (1 and n):

(1 + n)

Now how many PAIRS are there? Well that's easy, take half of the total number of numbers (sorry no other way I can see to say it)...there are a total of n numbers, so it's just half of that:

(n + 1) * n/2
--> or

n(n + 1)/2

...so after that, it's just a divide by n, so divide by n...then they cancel:

{ n(n + 1)/2 } / n
--> n's cancel, leaving

(n + 1) / 2

sum of 1 + 2 + 3 = middle number × total numbers
middle number = (1st number + last number)/2 = (1+3)/2 = 2
= 2 × 3 = 6
similarly middle number of 1 + 2 + 3 + ... + n = (n + 1)/2
so sum of 1 + 2 + 3 + ... + n = n(n + 1)/2
so n(n + 1)/2/n = n(n + 1)/2n
= (n + 1)/2
----

S=1+2....+n
2s=(n+1)+(n-1+2)+(n-2+3)...n times so s=n(n+1)/2