Solving Math Induction: Prove 1^3 + 2^3 + 3^3 + 4^3 ... + n^3 = ((n^2 + n)/2)^3

[ruud]

I'm stuck and not sure what I've done wrong for this problem

Prove the following by mathematical induction:
1^3 + 2^3 + 3^3 + 4^3 ... + n^3 = ((n^2 + n)/2)^3

ok so I proved it for n = 1 and n = 2 then assume n = k
so ((k^2 + k)/2)^3

Then let's try to do n = k +1
so
((k+1)^2 + k +1)/2)^3 = ((k^2 + k)/2)^3 + (k + 1)^3
after expanding I get

(k^6+ 9k^5+ 33k^4+ 63k^3+ 66k^2+ 36k+ 8)/8
=
(k^6 +3k^5 + 3k^4 + 9k^3 + 24k^2 + 24k + 8)/8 + k^3 + 3k^2 + 3k +1

For some reason I think that I don't have to do all of this expanding. Can someone please tell me what I"m doing wrong or what I need to fix?

 

[ruud]

Nevermind I just heard from a fellow friend that this statement is false.

 

[HallsofIvy]

Someone had to tell you that? Did you even try calculating a few numbers?

If n= 2, 1²+ 2²= 1+ 8= 9

$$\(\frac{n^2+n}{2}\)^3= \(\frac{4+2}{2}\)^3= 3^3= 27$$

so for the very second number its not true.

If you had done even the slightest amount of work on this you would have seen:
1 1³ = 1
2 1³+2³= 1+ 8= 9
3 1³+2³+3³= 1+ 8+ 27= 36
4 1³+2³+3³+4³= 100.

Hmmm: 1, 9, 36, 100,... what does that make you think of? Not cubes certainly!