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I've seen these two different geometric sum formulas but I don't know when which is used.

$S = a\left(\frac{1-(r^n)}{1-r}\right)$

and

$S = a\left(\frac{1-(r^{n+1})}{1-r}\right)$

André Nicolas
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Nwqp
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  • For n terms the first is used. If there are n+1 terms, (eg. You dont count the first term as a term) the second is used. – Kartik Oct 01 '15 at 02:07

1 Answers1

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The second one is the sum for one additional term.

Note that

$\begin{array}\\ a\left(\frac{1-(r^{n+1})}{1-r}\right)-a\left(\frac{1-(r^{n})}{1-r}\right) &=a\left(\frac{(1-(r^{n+1}))-(1-(r^{n}))}{1-r}\right)\\ &=a\left(\frac{r^n-r^{n+1}}{1-r}\right)\\ &=a\left(\frac{r^n(1-r)}{1-r}\right)\\ &=ar^n \end{array} $

and this is the next term that is added to the first sum to get the second sum.

marty cohen
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