Proof of sum of an arithmetics series

Sn=n/2[2a+(n-1)d]
why?

I don't know about proofing, but you can derive the formula from known statements and a cute thing where you add the series to the reverse order of itself to produce n terms that are all the same and then divide by 2.  Notice that each successive term is d more than the previous, and the n^th term would have (n - 1) of the d amount since the first term has zero d, the second has one d, the third has two d, etc., so that each term has one less d than its order number in the list.

S_n = a + (a + d) + (a + 2d) + ... + (a + (n-2)d) + (a + (n-1)d)
S_n = (a + (n-1)d) + (a + (n-2)d)+ ... + (a + 2d) + (a + d) + a

2S_n = (2a (n-1)d) + (2a (n-1)d) + (2a (n-1)d) + ... + (2a (n-1)d)
= n(2a (n-1)d)

S_n = (n(2a (n-1)d))/2 which is equivalent to your forumula.

fixing typo, last three lines are missing plus signs in every bracketed term, such as (n(2a + (n-1)d))/2

well done!
are you  teacher? :L
you explained it like a teacher
i award you a square slice of 1.772 :D

>>4
I make some money explaining this level of stuff for a few years, and contracted as adult instructor sometimes, so I got better at remembering it

>>5
do you last longer nwow?

my boners?  I take pills that are supposed to have a side effect to diminish them but I still have boners all night and it's getting annoying