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OpenStudy (anonymous):

Find all values of x for which Σ n=1to ∞( (x-2)/3)^n is a convergent geometric series, then express the sum of the series as a function of x.

OpenStudy (anonymous):

Looks horrible... :3

OpenStudy (anonymous):

calc 2 baby hahah

OpenStudy (anonymous):

Thankfully, I'm here to watch you :)

OpenStudy (anonymous):

Here, I'll even write it for you \[\huge \sum_{n=1}^\infty\left(\frac{x-2}3\right)^n \]

OpenStudy (anonymous):

Now, when is a geometric series convergent ^.^

OpenStudy (anonymous):

infininity over infinity

OpenStudy (anonymous):

?

OpenStudy (anonymous):

No, you misunderstand... I just asked when a geometric series is convergent...?

OpenStudy (anonymous):

If |r| < 1

OpenStudy (anonymous):

That's right :) Now what is r in this specific series?

OpenStudy (anonymous):

(x-2)/3?

OpenStudy (anonymous):

Right again :D So you want \[\huge \left|\frac{x-2}{3}\right|<1\]Now solve :>

OpenStudy (anonymous):

x<5

OpenStudy (anonymous):

and?

OpenStudy (anonymous):

x>-1

OpenStudy (anonymous):

Correct :) So your range for x is just (-1 , 5)

OpenStudy (anonymous):

now how can i express is at a sum of the series? do i do that a/1-r

OpenStudy (anonymous):

Yeah, and as usual, replace r properly... \[\huge r = \frac{x-2}{3}\]

OpenStudy (anonymous):

so 1/1-((x-2)/3)

OpenStudy (anonymous):

Yup, and simplify as desired ^.^

OpenStudy (anonymous):

then what

OpenStudy (anonymous):

simplify it, and I'll tell you what then :)

OpenStudy (anonymous):

3/1-x

OpenStudy (anonymous):

And that's actually it :)

OpenStudy (anonymous):

You have just expressed the sum of the series as a function of x.

OpenStudy (anonymous):

wow thank you so much you were a huge help :) you know your stuff

OpenStudy (anonymous):

That I do ^_^

OpenStudy (anonymous):

prof or past calc 2 student?

OpenStudy (anonymous):

Uhh.. neither :)

OpenStudy (anonymous):

10th grader, actually ^_^ (You just had to ask :> )

OpenStudy (anonymous):

no way

OpenStudy (anonymous):

your one smart 10th grader

OpenStudy (anonymous):

Well, I guess 10th grader is more believable than "ageless kid from the second to the right and straight on till morning" But I am in 10th grade, though :)

OpenStudy (anonymous):

how d you know calc two stuff

OpenStudy (anonymous):

Books. Too much time on my hands :)

OpenStudy (anonymous):

dang well thanks

OpenStudy (anonymous):

Any time (well, not really :> )

OpenStudy (anonymous):

^^^ this guy

OpenStudy (anonymous):

Cheerio ^.^

OpenStudy (anonymous):

so you were not completely right but i fixed the problem

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