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- anonymous

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

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- anonymous

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

- jamiebookeater

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- anonymous

\[\sum_{k=1}^{5} k^3\]

- amistre64

i think its just the square of k^1

- 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

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

Just to get a feel for it, you feel me!? ^^

- anonymous

Yeah, thanks. So would it be 1, 8, 27, 64, 125?

- across

Don't forget that it's a summation! :P

- anonymous

What do you mean?

- TuringTest

you have to add them...

- 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

euler plagarized the chinese :P

- anonymous

^^ lol, how do you know ? :P

- amistre64

past life regression :)

- amistre64

if we take the derivative of this life ....

- anonymous

You were there when Euler did it? ... omG! :D

- amistre64

yep, just call me JC Superstar yay!!

- anonymous

lol :D

- 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

ADD them up!
that is what summation means

- TuringTest

I said it above but nobody cared...

- TuringTest

\[1^3+2^3+3^3+4^3+5^3=?\]

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