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

#Calculus

Easy Math Editor

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Comments

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 - 3 years, 4 months ago

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

Poca Poca - 3 years, 4 months ago

Great work!

Pi Han Goh - 3 years, 4 months ago

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