# 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 Hakan Eskici
3 years, 6 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)

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Great work!

- 3 years, 6 months ago

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

- 3 years, 6 months ago

But does it work for $x = 0$ or for $x=-1$?

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

no, you can't signs must be altering like in the picture

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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 ?

- 3 years, 6 months ago

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

- 3 years, 6 months ago

hmm, what about if $r > 0$ normal geometric series formula cannot provide it

- 3 years, 6 months ago

It works too.

- 3 years, 6 months ago

What should I do now can you give advice?

- 3 years, 6 months ago

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 $S - Sr$?

- 3 years, 6 months ago

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.

- 3 years, 6 months ago