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

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\[\sum_{k=1}^{5} k^3\]
i think its just the square of k^1
\[\sum k=\frac{n(n+1)}{2}\] \[\sum k^3=\left(\frac{n(n+1)}{2}\right)^2\] rings a bell to me

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Other answers:

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

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