1. ## Formula

Can someone tell me the formula for finding the sum of a finite qeometric series

2. let $S=\sum\limits_{i=1}^{n}ar^{i-1}$

then
$S=a+ar+ar^{2}+\cdots +ar^{n-1}$

and

$rS=ar+ar^{2}+ar^{3}+\cdots +ar^{n}$

substract this two equations yield

$S(1-r)=a-ar^{n}$

$S=\frac{a-ar^{n}}{(1-r)}$

3. Originally Posted by dedust
let $S=\sum\limits_{i=1}^{n}ar^{i-1}$

then
$S=a+ar+ar^{2}+\cdots +ar^{n-1}$

and

$rS=ar+ar^{2}+ar^{3}+\cdots +ar^{n}$

substract this two equations yield

$S(1-r)=a-ar^{n}$

$S=\frac{a-ar^{n}}{(1-r)}$

How do u find n

4. Originally Posted by Tinhorn
How do u find n
HI

Consider this series : 1+2+3+4

the n here would be 4

is this what you want toknow ?

5. thank you

HI

Consider this series : 1+2+3+4

the n here would be 4

is this what you want toknow ?

would that logic also worked for 1 + 2/3 + 4/9 +........ + 64/729

7. Originally Posted by Tinhorn
would that logic also worked for 1 + 2/3 + 4/9 +........ + 64/729
$n$ is the number of terms in the series.

what is n for this series?

8. Originally Posted by dedust
$n$ is the number of terms in the series.

what is n for this series?
That's what I'm trying to figure out

9. look at the formula,.
the first term $a=1$, the ratio $r=2/3$, and
the last term is $64/729$
so

$ar^{n-1}=64/729$
$\left(\frac{2}{3}\right)^{(n-1)}=\frac{64}{729}=\left(\frac{2}{3}\right)^6$
$n-1=6$
$n=7$