Consider the series
$s=(1-t)S \sum_{i=0}^k(1+r)^{k-i}$, where $S>0, 0
Simplification of simple (geometric) series
-2
$\begingroup$
sequences-and-series
arithmetic-progressions
geometric-progressions
1 Answers
1
$$\sum_{i=0}^k(1+r)^{k-i}=\sum_{i=0}^k(1+r)^i=\frac{1-(1+r)^{k+1}}{1-(1+r)}=\ldots$$
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0the exponent is $k+1$ instead of $k$ – 2017-01-04
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0@SalahFatima Thanks, edited. – 2017-01-04
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0Thanks and what happens if I have $s=\sum_{i=0}^k(1+r)^{k-i}S_i$, where $S_i$ is positive for each $i$? – 2017-01-04
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0@Kyuss I've no idea what happens then. It dependens completely on what $\;S_i\;$ is, and it may even be it won't have a closed form... – 2017-01-05