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Mathematics 13 Online
OpenStudy (anonymous):

find the sum

OpenStudy (anonymous):

\[\sum_{n=1}^{\infty} \frac{ 1+4^n }{ 8^n }\]

OpenStudy (xapproachesinfinity):

did you check the convergence

OpenStudy (anonymous):

how is the sum performed here? is it used with \[\frac{ a }{ 1-r }\]

OpenStudy (anonymous):

it converges yes

OpenStudy (xapproachesinfinity):

hmm i' thinking of doing a separation \[\frac{1+4^n}{8^n}=(\frac{1}{8})^n+(\frac{1}{2})^n\]

OpenStudy (anonymous):

yes, and that is what i did to find convergence... how do we proceed with finding the sum?

OpenStudy (xapproachesinfinity):

hmm there should the summation of (1/2)^n from 1 to k

OpenStudy (xapproachesinfinity):

it is like you mentioned above just check your notes how to do it

OpenStudy (xapproachesinfinity):

i honestly forgot how to do it

OpenStudy (anonymous):

thanks anyway

OpenStudy (xapproachesinfinity):

http://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%E2%8B%AF this the summation

OpenStudy (xapproachesinfinity):

as i said they used what you said above

OpenStudy (anonymous):

do i do this separately? r= 1/2 and then again r= 1/8? and then add those two together?

OpenStudy (xapproachesinfinity):

the other some i'm thinking of \[(\frac{1}{2})^{3n}\]

OpenStudy (xapproachesinfinity):

yes of course!

OpenStudy (xapproachesinfinity):

each one separately

OpenStudy (xapproachesinfinity):

sum not some* typo

OpenStudy (anonymous):

\[\frac{ \frac{ 1 }{ 2 } }{ 1-\frac{ 1 }{ 2 } }+ \frac{ \frac{ 1 }{ 8 } }{ 1-\frac{ 1 }{ 8 } }\]

OpenStudy (xapproachesinfinity):

they are really the same after looking from a different angle they are given the same result

OpenStudy (xapproachesinfinity):

no my friend

OpenStudy (xapproachesinfinity):

look what i did 1/8 can be written as (1/2)^3 so we get ((1/2)^n)^3

OpenStudy (xapproachesinfinity):

so it is really the same sum raised to the power 3

OpenStudy (xapproachesinfinity):

the sum would be 2 1+1^3=1+1=2

OpenStudy (anonymous):

it says the answer is 8/7... but it does not show me how they got that.... so i know 2 is not correct

OpenStudy (xapproachesinfinity):

if you did it the other way you get 8/7

OpenStudy (xapproachesinfinity):

the way you suggested

OpenStudy (anonymous):

\[\frac{ \frac{ 1 }{ 8 } }{ \frac{ 8 }{ 8 }-\frac{ 1 }{ 8 } }= \frac{ 1 }{ 8 }(\frac{ 7 }{ 8 })= \frac{ 7 }{ 64 }\]

OpenStudy (anonymous):

i did that backwards

OpenStudy (xapproachesinfinity):

hmm that is incorret should 8/7 (1/8)

OpenStudy (xapproachesinfinity):

you have to flip the fraction

OpenStudy (anonymous):

\[\frac{ 1 }{ 8 }(\frac{ 8 }{ 7 })= \frac{ 8 }{ 56 }\]

OpenStudy (anonymous):

lol oops

OpenStudy (xapproachesinfinity):

you should have 1/7 +1 =8/7

OpenStudy (xapproachesinfinity):

which is what you are looking for

OpenStudy (anonymous):

yep. thamks!

OpenStudy (anonymous):

thanks

OpenStudy (xapproachesinfinity):

no problem

OpenStudy (anonymous):

i always need to "see" it,..i donno y

OpenStudy (xapproachesinfinity):

I'm wondering why the other way i did went off the loop it should work!! something has to do with infinity i guess

OpenStudy (anonymous):

it sounded so "nice" too! it is because 1/2 became just 1 when we applied it to the sum formula.... and 1 to the power of anything is just 1

OpenStudy (anonymous):

thanks for being a trooper with me through this

OpenStudy (xapproachesinfinity):

it is the problem of infinity for sure 1^oo is somewhat problematic

OpenStudy (anonymous):

true!

OpenStudy (xapproachesinfinity):

oh well you solved anyway :)

OpenStudy (anonymous):

yep and im sure ill be back on with another

OpenStudy (xapproachesinfinity):

hehe feel free :)

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