anonymous
  • anonymous
Expand the series and evaluate: ∑ {5(on top), k=1(on bottom), k^3 on the right side}
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sum_{k=1}^{5} k^3\]
amistre64
  • amistre64
i think its just the square of k^1
amistre64
  • amistre64
\[\sum k=\frac{n(n+1)}{2}\] \[\sum k^3=\left(\frac{n(n+1)}{2}\right)^2\] rings a bell to me

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across
  • across
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
  • across
Just to get a feel for it, you feel me!? ^^
anonymous
  • anonymous
Yeah, thanks. So would it be 1, 8, 27, 64, 125?
across
  • across
Don't forget that it's a summation! :P
anonymous
  • anonymous
What do you mean?
TuringTest
  • TuringTest
you have to add them...
anonymous
  • anonymous
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
  • amistre64
euler plagarized the chinese :P
anonymous
  • anonymous
^^ lol, how do you know ? :P
amistre64
  • amistre64
past life regression :)
amistre64
  • amistre64
if we take the derivative of this life ....
anonymous
  • anonymous
You were there when Euler did it? ... omG! :D
amistre64
  • amistre64
yep, just call me JC Superstar yay!!
anonymous
  • anonymous
lol :D
anonymous
  • anonymous
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
  • TuringTest
ADD them up! that is what summation means
TuringTest
  • TuringTest
I said it above but nobody cared...
TuringTest
  • TuringTest
\[1^3+2^3+3^3+4^3+5^3=?\]

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