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At the last step of the provided solution to this problem the author transforms $$\frac{e^{t(n+1)}-1}{e^t-1}=(1+e^t+e^{2t}+...+e^{nt})$$ It is probably something simple, but I am having trouble figuring it out. I tried dividing the numerator by denominator :

$$ \frac{e^{t(n+1)}-1}{e^t-1}=e^{n+1}+e^{(n+1)/t}+e^{(n+1)t^2}+... $$

its not the same, can someone tell me what the author did?

Problem and answer

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It's just the formula for the geometric sum $$\sum_{k=0}^n x^k=\begin{cases}n+1&\text{if }x=1\\\frac{x^{n+1}-1}{x-1}&\text{if }x\ne 1\end{cases}$$ applied to $x=e^t$. The formula is typically proved by induction on $n\ge 0$ or like this.

Added: I can't quite tell what you've done exactly when you said "I've tried to divide the numerator by the denominator": it looks like fair amount of errors of algebra; it might thus be in your best interest to revise or hone that part of your knowledge.

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  • $\begingroup$ I tried to do long polynomial division $\endgroup$ – Frank May 22 '18 at 10:40
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$1+x+x^2+x^3+ . . . +x^{n-2}+x^{n-1}+ x^{n}=\frac{x^{n+1}-1}{x-1}$

$x= e^t$

$\frac{e^{t(n+1)}-1}{e^t-1}=1+e^t+e^{2t}+e^{3t}+e^{4t}+ . . .+e^{nt}$

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