jesusfreak
Expand the series and evaluate:
∑ {5(on top), k=1(on bottom), k^3 on the right side}
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jesusfreak
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\[\sum_{k=1}^{5} k^3\]
amistre64
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i think its just the square of k^1
amistre64
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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
across
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It's always a good idea to list out the elements of series this small:\[1^3+2^3+3^3+4^3+5^3.\]
across
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Just to get a feel for it, you feel me!? ^^
jesusfreak
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Yeah, thanks. So would it be 1, 8, 27, 64, 125?
across
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Don't forget that it's a summation! :P
jesusfreak
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What do you mean?
TuringTest
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you have to add them...
FoolForMath
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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!
amistre64
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euler plagarized the chinese :P
FoolForMath
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^^ lol, how do you know ? :P
amistre64
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past life regression :)
amistre64
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if we take the derivative of this life ....
FoolForMath
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You were there when Euler did it? ... omG! :D
amistre64
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yep, just call me JC Superstar yay!!
FoolForMath
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lol :D
jesusfreak
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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?
TuringTest
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ADD them up!
that is what summation means
TuringTest
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I said it above but nobody cared...
TuringTest
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\[1^3+2^3+3^3+4^3+5^3=?\]