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anonymous
 one year ago
How do i find the sum??
http://i.imgur.com/tBwQ0Fi.png
I know to represent it as a geometric series but I am unsure what to do with the ^n1
anonymous
 one year ago
How do i find the sum?? http://i.imgur.com/tBwQ0Fi.png I know to represent it as a geometric series but I am unsure what to do with the ^n1

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1We can use geometric series test, so first lets find the partial sum for \[\sum_{n=1}^{\infty} \frac{ (4)^{n1} }{ 5^n }\] so we let n = 1

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[a_1 = \frac{ (4)^{11} }{ 5^1 } = \frac{ (4)^0 }{ 5 } = \frac{ 1 }{ 5 }\] and we know if r<1 it is then convergent to find r we can use \[a_1r=a_2 \implies \frac{ 1 }{ 5 }r=\frac{ 4 }{ 25 } \implies r = \frac{ 20 }{ 25 } = \frac{ 4 }{ 5 }\] \[\frac{ a_1 }{ 1r } = \frac{ 1/5 }{ 1+4/5 } = \frac{ 1 }{ 9 }\] therefore it's convergent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh i see we manipulated a2/a1=r

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1That gives us our first term a_1

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Remember r is the common ratio here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh right n=1 in the sum so its our first term, i keep on thinking it starts at zero. Thanks
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