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Wherever I see the sum of a infinite geometric series with $|r|<1$ being derived the series always starts at $n = 0$, or $n = 1$, the basic form is

$$a + ar + ar^2 + ar^3 + ... $$

And the sum is $\frac{a}{1-r}$

Does that still apply for a geometric series that starts at say n = 101, so

$$ar^{100} + ar^{101} + ar^{102} +... $$

FisherDisinformation
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Arnold Doveman
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2 Answers2

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\begin{align*} ar^{100} + ar^{101} + ar^{102} + \text{ ... } &= r^{100}\big[a + ar + ar^{2} + \text{ ... } \big] \\ &= \dfrac{ar^{100}}{1-r} \end{align*} So factoring out $r^{100}$, yes you have the same form.

miradulo
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  • I had always learned the formula as 'the first term over $1-r$.' In that case you can start with any index. I suppose your answer illustrates why this is true. – John Molokach Aug 20 '15 at 10:12
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I remember this by something a friend told me once. The sum of an infinite geometric series is the first term over one minus the common ratio. It is easier to remember without a formula I find.

7zf
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  • Yeah, I didn't think of it until now, but the numerator will always be the first term in the sequence you choose to start at. – Anson Savage Apr 17 '18 at 23:42