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OpenStudy (anonymous):
k=1 sigma k tends to infinity 4^(k+1)/7^(k-1) ?
this is my qustion how can calculte the sum of this series..?
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OpenStudy (turingtest):
\[\sum_{k=1}^\infty{4^{k+1}\over7^{k-1}}\]correct?
OpenStudy (anonymous):
yup.:)
OpenStudy (turingtest):
you need the actual sum, not just to know that it converges?
OpenStudy (anonymous):
yup i find that converges or not..? but how.?
OpenStudy (turingtest):
\[\sum_{k=1}^\infty{4^{k+1}\over7^{k-1}}=\sum_{k=1}^\infty16(\frac47)^{k-1}\]now it's a simple geometric series
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OpenStudy (turingtest):
the sum of any geometric series of the form\[\sum_{k=1}^\infty ar^{k-1}=\frac a{1-r}~~~~if~~~~~|x|<1\]
OpenStudy (anonymous):
bro can u explain ..how..?
OpenStudy (anonymous):
16 sigma ..? how..?
OpenStudy (turingtest):
\[4^{k+1}=4^2\cdot4^{k-1}\]
OpenStudy (turingtest):
\[=16(4^{k-1})\]
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OpenStudy (anonymous):
oho ok ok i understand bro
OpenStudy (turingtest):
got it from here?
OpenStudy (anonymous):
yes |r|<1 and a=16 me right.. na..?
OpenStudy (turingtest):
yep
OpenStudy (turingtest):
so the sum is...?
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OpenStudy (anonymous):
i calciule just 2 mints
OpenStudy (anonymous):
16/21 ?
OpenStudy (turingtest):
not what I got
OpenStudy (turingtest):
what is r ?
OpenStudy (anonymous):
r=4/7
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OpenStudy (turingtest):
so what is 1-r ?
OpenStudy (anonymous):
3/7
OpenStudy (turingtest):
so what is
16/(3/7) ?
OpenStudy (anonymous):
112/3
OpenStudy (turingtest):
there we go :)
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OpenStudy (anonymous):
thanks..:)
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