# Geometric series

From Maths

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## Contents

## Notes

For the geometric progression [ilmath](1,r)[/ilmath] we have:

- [math]S_n:\eq\sum^n_{k\eq 1}r^{k-1}\eq\frac{1-r^n}{1-r}[/math] (with first term [ilmath]n\eq 1[/ilmath], as is our convention (see:
*sequence*))

Thus for a general geometric progression, [ilmath](a,r)[/ilmath] we have:

- [math]S_n:\eq\sum^n_{k\eq 1}ar^{k-1}\eq a\sum^n_{k\eq 1}r^k{-1}\eq a\frac{1-r^n}{1-r}[/math]