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$ 3= \sum_{n=1}^{t} \frac{1}{1.08^n} $

I see that it is $3 = 1.08^{-t}(12.5 \times 1.08^t{-12.5})$ (from Wolfram Alpha, but I'm not sure how to get it. I tried solving as a geometric series, I had problems and didn't get the correct answer.

I see that $ t\approx3.56592$, which seems like it's correct, but I have no idea where the 12.5 and all that stuff came from. Unfortunately my calculus book doesn't help much, as it is mainly focus on infinite series.

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    @Americo: Exactly! The context is relevant, that is why I was asking OP for a different phrasing of the problem which will make it clearer (see my earlier comments). As the question is currently stated the only context I can meaningfully derive is that$t$is a natural number, and the fact that$S(t)$can be extended to reals/complex is kind of irrelevant... You might have interpreted it different, but I find that it is better to let OP clarify rather than we try to guess what OP might have meant.2010-10-07

3 Answers 3

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I would solve for $t$

$S(t)=\displaystyle\sum_{n=1}^{t}\dfrac{1}{1.08^{n}}=3.$

The sum of a geometric progression with first term $u_{1}$ and ratio $r$ is given $(\ast)$ by

$S(t)=u_{1}\times \dfrac{1-r^{t}}{1-r}.$

In this case

$\dfrac{1}{1.08}\times \dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{1-\dfrac{1}{% 1.08}}=3$

or

$\dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{0.08}=3\iff \left( \dfrac{1}{1.08}% \right) ^{t}=0.76.$

Applying logarithms, we obtain the "time" $t$

$t\log \left( \dfrac{1}{1.08}\right) =\log 0.76,$

$t=\dfrac{\log 0.76}{-\log 1.08}\approx 3.5659,$

which means that we need more than $3$ periods and less than $4$, at an interest rate of $8\%$, coumpounded per period, to get a total of $3$ currency units.

$(\ast)$ Derivation:

$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$

$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$

$u_{k}=ru_{k-1}=u_{1}r^{k-1}$

$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left( ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right) $

$(1-r)S=u_{1}-ru_{t}$

$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$

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HINT$\ $ Put $\rm\ a = 1.08\:$. Multiply both sides by $\rm\ a^t\ $ to get$\:$ RHS $\rm\: = \ \sum_{k=0}^{t-1}\ a^k$

As for "where the $12.5$ came from", $\rm\ 1/(a-1) = 1/0.08 = 100/8 = 12.5$

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Simply suppose that $r = \frac{1}{1.08}$, or $r = a^{-k} = \frac{1}{a^k}$ and solve in the usual manner, as you would if the exponent was positive.

Example

$ \sum \limits_{i=0}^{k-1} 9^{-i} = \sum \limits_{i=0}^{k-1} \frac{1}{9^i} = \frac{1- \left( 1/9 \right)^k}{1- \left( 1/9 \right)} = \frac{1 - \left( 1 / 9 \right)^k}{8/9} = \frac{9}{8} \left( 1 - \frac{1}{9^{k}} \right) $

(Do not be confused with the harmonic series which does not have an exponent in its term.)