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wcaprar

  • 2 years ago

What is the sum of the infinite series?

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  1. wcaprar
    • 2 years ago
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    \[\sum_{n=2}^{\infty} 2^n/3*5^{n+2}\]

  2. wcaprar
    • 2 years ago
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    I see that you can take out the 5^2 and multiply that with three then simplify to get \[(1/75)*(2/5)^n\] which is a geometric series. But when I solve it out like you would a regular geometric series I get 1/45. But wolframalpha says 20

  3. joemath314159
    • 2 years ago
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    your idea is correct. lets make sure the arithmetic is correct

  4. joemath314159
    • 2 years ago
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    \[\frac{1}{75}\sum_{n=2}^{\infty}\left(\frac{2}{5}\right)^n=\frac{1}{75}\left(\frac{\frac{4}{625}}{1-\frac{2}{5}}\right)\]

  5. wcaprar
    • 2 years ago
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    How do you get 4/625 on top?

  6. joemath314159
    • 2 years ago
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    we are starting at n=2, not n=1. The first term of the sequence is 4/625.

  7. wcaprar
    • 2 years ago
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    Oh! The first term!

  8. wcaprar
    • 2 years ago
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    Duh! Thanks!

  9. joemath314159
    • 2 years ago
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    yw! :)

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