geometic series of n terms

**kcsteven** · February 6th 2007, 09:22 AM

A geometric series of n terms can be wiriiten:
a + ar + ar^2 + ar^3 +... + ar^n-1 where a, is the first and r is the common ratio, how would you put this in sigma notation? How can a formula be written in a, r and n for the sum of this series?
Thank you!
Keith Stevens

**Plato** · February 6th 2007, 09:27 AM

Once again it is simple: $a\sum\limits_{k = 0}^{n - 1} {r^k }.$

**CaptainBlack** · February 6th 2007, 10:41 AM

Originally Posted by kcsteven

A geometric series of n terms can be wiriiten:
a + ar + ar^2 + ar^3 +... + ar^n-1 where a, is the first and r is the common ratio, how would you put this in sigma notation? How can a formula be written in a, r and n for the sum of this series?
Thank you!
Keith Stevens

Observe that if we multiply the series by $1-r$ we get all the terms but the first and last cancelling:

$<br /> \left[ \sum_{k=0}^n a\,r^k \right](1-r)=a(1-r^{n+1})<br />$

so:

$<br /> \sum_{k=0}^n a\,r^k =a\,\frac{1-r^{n+1}}{1-r}<br />$

**Soroban** · February 6th 2007, 11:12 AM

Hello, Keith!

Here's where the summation formula comes from . . .

We have the series: . $S \;=\;a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}$

. . . . Multiply by $r\!:\;rS \;=\;\qquad ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n$

Subtract and we get: . $S - rS \;=\;a - ar^n$

Therefore: . $(1 - r)S \;=\;a(1- r^n)\quad\Rightarrow\quad \boxed{S \;=\;a\,\frac{1-r^n}{1 - r}}$

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