geometric series...need major help.

**Steph07** · December 3rd 2008, 06:29 PM

Find the sum of the following infinite
geometric series. Use a formula and
show your work.

(100)+(-40)+(16)+(-32)+etc...forever
5
Hint: r = (-2)
5

thats -32/5.... and on the hint is -2/5

**terr13** · December 3rd 2008, 06:41 PM

Do you mean its - $32/5$ ?

Since the hint says to try $r =-2/5$ , we can plug that in to our values and see what happens
$100(-2/5) = -40$
$ -40(-2/5) = 16 $

So we see that each term is changing by a factor of $-2/5$ . Now we know the equation for a geometric series is $a/(1-r),$ where a is the first term and r is the ratio. Plug in the first term in the series for a, and plug in the ratio r, and then solve the equation to find the sum of the series.

**vincisonfire** · December 3rd 2008, 06:42 PM

$\sum_{k=0}^{n-1} ar^k=a\frac{1-r^n}{1-r}$
When $\lim_{n\rightarrow \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r}$ since r < 1

**Steph07** · December 3rd 2008, 06:46 PM

so how would i solve that, I dont even know what the E thing is...

**Soroban** · December 3rd 2008, 06:48 PM

Hello, Steph07!

Find the sum of the following infinite geometric series:
. . $100 -40 + 16 - \tfrac{32}{5} + \hdots$

You're expected to know the formula . . .

. . $S \;=\;\frac{a}{1-r}$ . . . where $a$ is the first term and $r$ is the common ratio.

We have: . $a = 100,\;r = \text{-}\tfrac{2}{5}$

Plug them into the formula . . .

**terr13** · December 3rd 2008, 06:51 PM

$ \sum $
means the summation, but I guess you don't really need to know that, just plug into the formula will get you your answer.

**Steph07** · December 3rd 2008, 07:01 PM

so is the answer 100.4 ?

**Steph07** · December 3rd 2008, 07:07 PM

or is it 71.4 ?

**terr13** · December 3rd 2008, 07:11 PM

I think it should be 71.4.

$ 100/(1-(-2/5)) = 100/(1+2/5) = (100/(7/5)) = 500/7 $

**Steph07** · December 3rd 2008, 07:13 PM

oOo... I see now... thank u

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