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

1. jesusfreak

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

2. amistre64

i think its just the square of k^1

3. amistre64

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

4. across

It's always a good idea to list out the elements of series this small:$1^3+2^3+3^3+4^3+5^3.$

5. across

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

6. jesusfreak

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

7. across

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

8. jesusfreak

What do you mean?

9. TuringTest

10. FoolForMath

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!

11. amistre64

euler plagarized the chinese :P

12. FoolForMath

^^ lol, how do you know ? :P

13. amistre64

past life regression :)

14. amistre64

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

15. FoolForMath

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

16. amistre64

yep, just call me JC Superstar yay!!

17. FoolForMath

lol :D

18. jesusfreak

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?

19. TuringTest

ADD them up! that is what summation means

20. TuringTest

I said it above but nobody cared...

21. TuringTest

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