Mathematics
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OpenStudy (anonymous):
find the sum
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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
\[\frac{1+4^n}{8^n}=(\frac{1}{8})^n+(\frac{1}{2})^n\]
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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
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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!
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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
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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
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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
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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!
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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
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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
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OpenStudy (xapproachesinfinity):
hehe feel free :)