Un to Sn in arith. seq.

[Acidic]

I was bored today and so I was wondering how one can convert U_n to S_n 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 S_n is the area under the curve.

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

I thought that S_n = 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 U_n, 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 U_n=-2n+8
then
S_n=-n²+8n-n
=-n²+7n

where on the graph doesn the -n fit in?

[twanvl]

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

You are sugesting Int(U_n,n)+1/2an. Where
S_n = U_n = an + b
Int(U_n,n)-n = 1/2an² + bn + 1/2an
My formula gives:
S_n = 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... )

[Acidic]

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

that should give the 0th term.