Un to Sn in arith. seq.

[Acidic]

I was bored today and so I was wondering how one can convert Un to Sn without having to find the terms and use crappy formula, I wanted to find a straight one-step calculation.

I thought, logically, if you draw:
y=n

then Sn is the area under the curve.

I then sketched:
y=2n+1
y=3n+1
y=4n+1

I thought that Sn = area under triangle (n2/2 (for y=2n+1))+area of triangle (nU1)

Anyway, to cut a boring story short, I found that you integrate Un, then add on half of the coeff of n.

the adding on at the end was an accidental find, just pot luck that it worked.

Now I cannot see where on the graph that bit fits in. Can you help me out?

I'll give an example of how my theory works.
you have Un=-2n+8
then
Sn=-n2+8n-n
=-n2+7n

where on the graph doesn the -n fit in?

[twanvl]

The formula I learned is
Sn = 1/2 * n * (U1 + Un)
The sum of all terms = n * the mean of the terms = n * the mean of the first and last term
In your example:
Un=-2n+8
then
Sn=1/2 * n * (-2+8 + -2n+8)
= 1/2 * n * (14 - 2n)
=7n - n2

You are sugesting Int(Un,n)+1/2an. Where
Sn = Un = an + b
Int(Un,n)-n = 1/2an2 + bn + 1/2an
My formula gives:
Sn = 1/2 * n * (a+b + an+b) = 1/2an2 + bn + 1/2an
You can see that these formulas are the same. The reason you need the extra term is because the first term of the sequence is term 1, not term 0. If you want the sum of terms 0...n you need the extra term

(Damn, that's a lot of terms, I probably got the term(inology) for the terms of the sequence wrong... :D)

[Acidic]

ok, but why can't you add on U1-d where d is the common difference?

that should give the 0th term.