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Jonask

  • 3 years ago

sigma

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  1. ShotGunGirl
    • 3 years ago
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    What?

  2. Jonask
    • 3 years ago
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    \[\sum_{k=1}^{49}(k+1)^3-k^3\] evaluate

  3. Jonask
    • 3 years ago
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    no formular jus a series

  4. J_Liu
    • 3 years ago
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    That series is just saying when you plug in k = 1, 2, 3, 4....49 and add them up, there should be a formula for the series to make the calculation easier. It is possible to calculate every term.

  5. Jonask
    • 3 years ago
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    \[2^3-1^3+3^3-2^3+4^3-3^3+...+50^3-49^3\]

  6. TuringTest
    • 3 years ago
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    it appears to be a telescoping series

  7. Jonask
    • 3 years ago
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    does it mean numbers are cancelling out

  8. Jonask
    • 3 years ago
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    i think the anwers is \[-1^3+50^3\]

  9. amistre64
    • 3 years ago
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    \[\sum_{k=1}^{49}(k+1)^3-k^3\] \[\sum_{k=1}^{49}(k+1)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=1+1}^{49+1}(k)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=2}^{50}k^3-\sum_{k=1}^{49}k^3=50^3-1^3\]

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