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I get how the formula is made. What I don't get is why whoever decided to multiply the equation by $(1-r)$. I'm referencing this link at Wikipedia. I get the mechanics, just not some of the logic.

Consider $$ \sum_{i=0}^{n-1} a^i=a^0+a^1+a^2+...+a^{n-2}+a^{n-1} $$

I get that the ratio is $r$, but why multiply this equation by $(1-r)$ and not some other quantity minus $r$?

Thanks \ AndyAndy

Yuxiao Xie
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Andrew Falanga
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2 Answers2

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Actually the author is skipping a step. First you multiply the series by $r$ and then you subtract this from the original series to get that:

$$\sum_{k=1}^{n} ar^{k-1} - r\sum_{k=1}^{n} ar^{k-1} = (1-r)\sum_{k=1}^{n} ar^{k-1}$$

But expanding the left hand side we can see that lots of cancelation occurs.

Stefan4024
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Actually, it goes the other way and comes from a high school formula: \begin{align*} 1-a^2 &=(1-a)(1+a),\\ 1-a^3 &=(1-a)(1+a+a^2),\\ \vdots\\ 1-a^n &=(1-a)(1+a+a^2+\dots+a^{n-1}). \end{align*}

Bernard
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