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

    (brief is enough)
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    (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.
    • Thread Starter
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    using n ---> infinity?
    • Thread Starter
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    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?
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    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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