Does my induction proof for sum of cubes formula work?

[JasonRox]

I am writing the exact problem.

1. Prove that for all natural numbers n the following equality is satisfied:

$$1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}$$

I am doing the exact method they used in the book, which I don't entirely agree on.

My answer is different than the one in the back of the book, but they come to the same conclusion.

Let n = k + 1,

$$\frac{(k+1)^2(k+1+1)^2}{4} = \frac{k^4+6k^3+13k^3+12k+4}{4} = \frac{k^2(k+1)^2}{4} + (k+1)^3$$

Since, $\frac{k^2(k+1)^2}{4}$ is true by our assumption then adding $(k+1)^3$ does not change the solution when n = k + 1.

I left some of the calculations out so I don't have to waste an hour doing tex, but I don't think there is any mistakes in it.

Obviously, if you test it, it is right.

Does that constitute as a proof?

I don't agree with the way they do it because they are assuming that there assumption is true. I really don't know.

Any comments?

 

[arildno]

So, what you're objecting to is their expansion of what ought to be the (k+1)-sum?

 

[arildno]

I can't really agrree to the objection:
They are using the "k-true" assumption on their right-hand side of an equality.
By noting therefore that the right-hand side is nothing but the k+1-sum, they've proven the induction step in an admittedly roundabout, but, IMO, valid way.

There is NO assumption lying in expanding the ("arbitrarily chosen" ) expression on the left-hand side!

 

[Gokul43201]

JasonRox, do you know how inductive proofs work ?

 

[robert Ihnot]

The problem here, I feel, is that Jason Rox left out a very, very important fact: he failed to consider proving the formula for a small value, here k=1. I think that is why he does not see the induction. For k=1, the formula gives:

$$1^3=1=\frac{1^2*2^2}{4}$$

If it is true for k it is also true for k+1, and it is shown true for 1; thus since it is true for k=1, so it is also true for k+1 = 2, and if it is true for k=2 then it is true for k=3, and so on and so on.

 

[JasonRox]

Sorry, I forgot to mention that it works for k=1.

I understand why that is necessary. If it didn't work for one, then it wouldn't work for all natural numbers u_n if u_1 + u_2 + ... + u_n.

Note for Gokul: I have no clue how inductive proofs work. I'll make a note to read about it in one of my books.

 

[devious_]

Note for Gokul: I have no clue how inductive proofs work. I'll make a note to read about it in one of my books.

If it is true for k it is also true for k+1, and it is shown true for 1; thus since it is true for k=1, so it is also true for k+1 = 2, and if it is true for k=2 then it is true for k=3, and so on and so on.

...

 

[JasonRox]

...

I thought about it in that fashion, but I just didn't think it was enough.

It makes sense if you think about it. I haven't done proofs in this fashion before.