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jesusfreak

  • 3 years ago

Expand the series and evaluate: ∑ {5(on top), k=1(on bottom), k^3 on the right side}

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  1. jesusfreak
    • 3 years ago
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    \[\sum_{k=1}^{5} k^3\]

  2. amistre64
    • 3 years ago
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    i think its just the square of k^1

  3. amistre64
    • 3 years ago
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    \[\sum k=\frac{n(n+1)}{2}\] \[\sum k^3=\left(\frac{n(n+1)}{2}\right)^2\] rings a bell to me

  4. across
    • 3 years ago
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    It's always a good idea to list out the elements of series this small:\[1^3+2^3+3^3+4^3+5^3.\]

  5. across
    • 3 years ago
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    Just to get a feel for it, you feel me!? ^^

  6. jesusfreak
    • 3 years ago
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    Yeah, thanks. So would it be 1, 8, 27, 64, 125?

  7. across
    • 3 years ago
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    Don't forget that it's a summation! :P

  8. jesusfreak
    • 3 years ago
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    What do you mean?

  9. TuringTest
    • 3 years ago
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    you have to add them...

  10. FoolForMath
    • 3 years ago
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    Deriving this proof is a good excercise, there are many ways actually, but I like way of differentiating and then substituting .. probably that's how euler did it!

  11. amistre64
    • 3 years ago
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    euler plagarized the chinese :P

  12. FoolForMath
    • 3 years ago
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    ^^ lol, how do you know ? :P

  13. amistre64
    • 3 years ago
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    past life regression :)

  14. amistre64
    • 3 years ago
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    if we take the derivative of this life ....

  15. FoolForMath
    • 3 years ago
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    You were there when Euler did it? ... omG! :D

  16. amistre64
    • 3 years ago
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    yep, just call me JC Superstar yay!!

  17. FoolForMath
    • 3 years ago
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    lol :D

  18. jesusfreak
    • 3 years ago
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    Okay, I hope you two know that that wasn't very helpful. I am just wondering how to solve this. Across, what do you mean by a summation?

  19. TuringTest
    • 3 years ago
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    ADD them up! that is what summation means

  20. TuringTest
    • 3 years ago
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    I said it above but nobody cared...

  21. TuringTest
    • 3 years ago
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    \[1^3+2^3+3^3+4^3+5^3=?\]

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