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Mathematics 26 Online
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..?

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

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):

\[=16(4^{k-1})\]

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...?

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

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 :)

OpenStudy (anonymous):

thanks..:)

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