New formula about Geometric series with altering signs

I don't know maybe someone worked with such geometric series and created formulas but although I researched so much on the web I haven't seen anything.If anyone interested in I can send algebraic proofs and we can work together.Since I am not a math expert I cannot proof further such as specific number set but I think the ''X'' can be in real numbers set I tried a lot of numbers also I used calculators It is working correctly

Note by Haakon Natursson
7 months ago

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Proof: \[S = 1+x+x^2+...+x^n\] \[S \times x = x+x^2+...+x^{n+1}\] \[S - S \times x = S(1-x) = 1 - x^{n+1}\] \[S = \frac{1 - x^{n+1}}{1-x} \] You can choose x < 0; also generally \[(-1)^k \times a^k = (-a)^k\] Because I can see negative signs in your solution I assume that you want to choose x >= 0. \[S=\frac{1 - x^{n+1}}{1-x} \] \[S'=\frac{1 - (-x)^{n+1}}{1-(-x)}\] \[S'=\frac{1 - (-x)^{n+1}}{1+x}\] Now we multiply the denominator and the enumerator by x and get: \[S'=\frac{x + (-x)^{n+2}}{x+x^2}\] Q.E.D. (Note that the - changes to a + because the exponent of the -1 is raised by 1)

Poca Poca - 7 months ago

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Thank you so much.You approached with more elegant way.

Haakon Natursson - 6 months, 4 weeks ago

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Note that for x=0; x=+/-1 the series is not a geometric series.

Poca Poca - 7 months ago

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Great work!

Pi Han Goh - 7 months ago

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But does it work for \(x = 0 \) or for \(x=-1\)?

Pi Han Goh - 7 months ago

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No, I don't think so until now only they are the exceptions.

Haakon Natursson - 7 months ago

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In general, can you find a simple expression of \(a + ar + ar^2 + \cdots + ar^{n-1} \)?

Pi Han Goh - 7 months ago

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@Pi Han Goh no, you can't signs must be altering like in the picture

Haakon Natursson - 7 months ago

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@Haakon Natursson Are you saying that if \(r < 0 \), then we can't find a simple expression of \(a + ar + \cdots + ar^{n-1} \)?

Pi Han Goh - 7 months ago

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@Pi Han Goh oh sorry, you see why I am asking for help :D,it works with negative numbers even with irrational or rational ones.Why you aren't testing ?

Haakon Natursson - 7 months ago

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@Haakon Natursson I'm trying to tell you that all your equations came from the same formula.

Pi Han Goh - 7 months ago

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@Pi Han Goh hmm, what about if \(r > 0\) normal geometric series formula cannot provide it

Haakon Natursson - 7 months ago

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@Haakon Natursson It works too.

Pi Han Goh - 7 months ago

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@Pi Han Goh What should I do now can you give advice?

Haakon Natursson - 7 months ago

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@Haakon Natursson In general, can you find a simple expression of \(a + ar + ar^2 + \cdots + ar^{n-1} \)?

Hint: Let \(S\) denote the value of this expression. What is \(S - Sr \)?

Pi Han Goh - 7 months ago

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@Pi Han Goh yes we can with this formula [a(1 - rn )/ 1 - r].Have you been asking me this for the whole time? I thought we were talking about my formula.Okay, I am listening.

Haakon Natursson - 7 months ago

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