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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 ^n-1

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  1. Astrophysics
    • one year ago
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    We can use geometric series test, so first lets find the partial sum for \[\sum_{n=1}^{\infty} \frac{ (-4)^{n-1} }{ 5^n }\] so we let n = 1

  2. Astrophysics
    • one year ago
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    \[a_1 = \frac{ (-4)^{1-1} }{ 5^1 } = \frac{ (-4)^0 }{ 5 } = \frac{ 1 }{ 5 }\] and we know if |r|<1 it is then convergent to find r we can use \[a_1r=a_2 \implies \frac{ 1 }{ 5 }r=-\frac{ 4 }{ 25 } \implies r = -\frac{ 20 }{ 25 } = -\frac{ 4 }{ 5 }\] \[\frac{ a_1 }{ 1-r } = \frac{ 1/5 }{ 1+4/5 } = \frac{ 1 }{ 9 }\] therefore it's convergent

  3. Astrophysics
    • one year ago
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    Your sum is 1/9

  4. anonymous
    • one year ago
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    Oh i see we manipulated a2/a1=r

  5. Astrophysics
    • one year ago
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    Yup

  6. anonymous
    • one year ago
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    Why did we use n=1?

  7. Astrophysics
    • one year ago
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    That gives us our first term a_1

  8. Astrophysics
    • one year ago
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    Remember r is the common ratio here

  9. anonymous
    • one year ago
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    Oh right n=1 in the sum so its our first term, i keep on thinking it starts at zero. Thanks

  10. Astrophysics
    • one year ago
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    Np :)

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