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1. Can anyone tell me how to proof:
Sum of a Arithmetic Progression
Sum of a Geometric Progression
Sum to Infinity

(brief is enough)
2. (Original post by keisiuho)
Can anyone tell me how to proof:
Sum of a Arithmetic Progression
Sum of a Geometric Progression
Sum to Infinity

(brief is enough)
AP:

Sn = a + (a + d) + (a + 2d) + ... + [a + (n - 1)d]

Write backwards:

Sn = [a + (n - 1)d] + [a + (n - 2)d] + ... + a

Add: 2Sn = [2a + (n - 1)d] + [2a + (n - 1)d] + [2a + (n - 1)d] + ...

Clearly, 2Sn = n[2a + (n - 1)d]

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

GP:

Sn(x) = 1 + x + x^2 + x^3 + ... + x^n

Multiply by x: xSn(x) = x + x^2 + x^3 + ... + x^n + x^(n + 1)

Subtract: Sn(x) - xSn(x) = 1 - x^(n + 1) (all the terms in between cancel)

=> Sn(x)(1 - x) = 1 - x^(n + 1)

=> Sn(x) = [1 - x^(n + 1)]/(1 - x)

Can't remember how to do sum to infinity.
3. using n ---> infinity?
4. I can think of this:
as |x|<1
when n is very large (tends to infinity), x^n+1 tends to zero
so it becomes 1/(1-x)

is this acceptable?
5. Sum to infinity of a geometric series is a(1-r^n)/(1-r), nylex has the right proof he's just written 1 too many terms because the first term is just a, so the nth term is ar^(n-1).

Keisiuho - that's fine

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