
Nice Proof of a Geometric Progression Sum - nreece
http://sputsoft.com/archives/25
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huhtenberg
For this proof to be complete, it needs to be shown that the trapezoid
sequence fills ABC triangle in its entirety. I.e. that for every point between
A and C there is a trapezoid in the sequence that covers it. Not hard to do by
any means, but the resulting proof is not going to be purely visual anymore.

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antiform
Why is this completely necessary? What would be an acceptable proof to you of
this statement?

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cousin_it
Neat illustration, but I'd present the proof to kids in the usual manner:

(1 - r)(1 + r + r^2 + ... + r^n) = 1 - r^(n + 1), because all other terms
cancel out.

If |r| < 1, the right part tends to 1, QED.

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losvedir
I can never be quite sure that I'm remembering the proper closed form
expression, so I always rederive it in a way that's basically what you have
here, but a little less dense:

Call the sum of the terms Sn, though we don't know what it is.

Sn = 1 + r +r^2 + . . . + r^n

Multiply each side by r

(Sn)(r) = r + r^2 + . . . + r^(n+1)

Subtract the former from the latter

Sn - (Sn)(r) = 1 - r^(n+1)

Sn(1 - r) = 1 - r^(n+1)

Sn = (1 - r^(n+1)) / (1 - r)

And there we have the closed form expression for Sn. As n goes to infinity, it
approaches 1 / (1-r)

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Locke1689
This is the correct way to prove the summation. Although, I would point out
that for r > 1, it's probably better to phrase it as ( r^(n+1) - 1)/(r-1).
However, it's important to note that this is only true from a series starting
from zero. For a series starting from m it is:

    
    
      (r^(n+1)-r^m)/(r-1)
    

Which is trivially deducible from your method or the parent's method. For some
reason I tend to dislike geometric proofs. This is probably because I find it
very easy to make proof fallacies visually, more so than symbolically.
However, this geometric proof is also less useful than the symbolic because it
is much harder to use it to prove the summation at values smaller than
infinity, whereas the symbolic works just as well for n.

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lrm242
I find this kind of post on HN very enjoyable. Thanks for sharing.

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Dilpil
Is this the reason it is referred to as a Geometric progression?

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antiform
Sort of. The way I was taught, a progression is called arithmetic because each
term is the arithmetic mean of its predecessor and successor, i.e. if A comes
before a term X and B comes after a term X in an arithmetic sequence, then X =
(A+B)/2. Similarly, a geometric progression uses the geometric mean, i.e. X =
(AB)^(1/2).

