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jesusfreak

  • 3 years ago

sum_(k=1)^6-3(2)^k-1

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  1. jesusfreak
    • 3 years ago
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    \[\sum_{k=1}^6{3(-2)^k-1}\] The -1 is up with the k. Find the indicated sum of the geometric series.

  2. satellite73
    • 3 years ago
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    \[\sum_{k=1}^6{3(-2)^{k-1}}\]

  3. jesusfreak
    • 3 years ago
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    Find the indicated sum of the geometric series.

  4. satellite73
    • 3 years ago
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    with so few terms i would just add them

  5. jesusfreak
    • 3 years ago
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    How so? And I messed up. It is -3(2). Sorry

  6. satellite73
    • 3 years ago
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    \[\sum_{k=1}^6{3(-2)^{k-1}}=3\sum_{k=1}^6{(-2)^{k-1}}\] \[=3(1-2+4-8+16-32)\]

  7. satellite73
    • 3 years ago
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    \[\sum_{k=1}^6-3(2)^{k-1}\]

  8. satellite73
    • 3 years ago
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    like that?

  9. jesusfreak
    • 3 years ago
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    yes

  10. satellite73
    • 3 years ago
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    \[\sum_{k=1}^6-3(2)^{k-1}=-3\sum_{k=1}^6(2)^{k-1}\]

  11. satellite73
    • 3 years ago
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    \[-3\sum_{k=1}^6(2)^{k-1}=-3(1+2+4+8+16+32)=-3(63)\] if my arithmetic is correct

  12. jesusfreak
    • 3 years ago
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    Yes! Thank you so much satellite73!!!! :)

  13. satellite73
    • 3 years ago
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    yw

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