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I was working on a mathematical stats problem and I don't get this part (that comes from a recursive arithmetico-geometric series):

$$U_{k} = \alpha C^{2} + \alpha^{2} C^{2} + \alpha^{3} C^{2} + ... +\alpha^{k-1} C^{2}+ \alpha ^{k} U_{0}$$

$$= \alpha C^{2} \left [ 1 + \alpha +\alpha ^{2}+ ... + \alpha ^{k-2} \right ] + \alpha ^{k}U_{0}$$

$$= \alpha C^{2} (\frac{\alpha ^{k-1}-1}{\alpha - 1}) + \alpha ^{k}U_{0}$$

My particular issue is how to go from the second line to the last one. I guess there is a closed form expression of that sum but I couldn't find it anywhere in that exact form.

Thanks.

Metrician
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2 Answers2

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There is an error after the second $=$ sign; it should read $\alpha C^2(1+\alpha+\alpha^2+\cdots+\alpha^{k-2})+\alpha_kU_0$. Now, use the fact that$$1+\alpha+\alpha^2+\cdots+\alpha^{k-2}=\frac{1-\alpha^{k-1}}{1-\alpha}.$$

José Carlos Santos
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The general form of a geometric series is $$a+ax+ax^2+\dots+ax^{n-1}=a(1+x+x^2+\dots+x^{n-1})=\frac{a(x^n-1)}{x-1}=\frac{a(1-x^n)}{1-x}$$

Peter Foreman
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  • Exactly. That last edit is what cleared my confusion. So I guess the two terms are interchangeable because any eventual negative sign would get cancelled out? – Metrician Apr 13 '19 at 13:22
  • Sorry, I don't know what you mean by an 'eventual negative sign' – Peter Foreman Apr 13 '19 at 13:23
  • The fact that these two expressions $\frac{a(x^n-1)}{x-1}=\frac{a(1-x^n)}{1-x}$ were equivalent was the source of my confusion. So I was asking if they were equivalent because the division would cancel out the negative sign resulting from the subtraction – Metrician Apr 13 '19 at 13:27
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    Yes because $\frac{a}{b}=\frac{-a}{-b}$ we can say that $\frac{x^n-1}{x-1}=\frac{-(x^n-1)}{-(x-1)}=\frac{1-x^n}{1-x}$ – Peter Foreman Apr 13 '19 at 13:28
  • That was my initial intuition but thought there was something deeper than that. I was reading too much into it. Thanks for your answer ! – Metrician Apr 13 '19 at 13:30