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

Question

anonymous · Answer

$$\sum_{k=1}^{5} k^3$$

amistre64 · Answer

i think its just the square of k^1

amistre64 · Answer

$$\sum k=\frac{n(n+1)}{2}$$
$$\sum k^3=\left(\frac{n(n+1)}{2}\right)^2$$
rings a bell to me

across · Answer

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 · Answer

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

anonymous · Answer

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

across · Answer

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

anonymous · Answer

What do you mean?

turingtest · Answer

you have to add them...

anonymous · Answer

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 · Answer

euler plagarized the chinese :P

anonymous · Answer

^^ lol, how do you know ? :P

amistre64 · Answer

past life regression :)

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

yep, just call me JC Superstar yay!!

anonymous · Answer

lol :D

anonymous · Answer

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 · Answer

ADD them up!
that is what summation means

turingtest · Answer

I said it above but nobody cared...

turingtest · Answer

$$1^3+2^3+3^3+4^3+5^3=?$$