Can You Prove the Sum of Cubes Equals the Square of the Sum?

[dleacock]

hey everyone,
I recently bought the the Dover Series book on Number Theory, and the 2nd example on page 5 asks your to prove
$$1^3 + 2^3 + 3^3 ... + n^3 = (1 + 2 + 3...)^2$$

Now, we've already proved that $$S_n = \frac{n(n+1)}{2}$$

So here's how I proved it...
$$(S_n)^2 = (\frac{n(n+1)}{2})^2$$

before we proved how $$S_k+1 = S_k + (k + 1)$$

Which lead me to...
$$\frac{1}(k+1)^2((k+1)+1)^2{4} + (k+1)^2$$
$$= \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2}$$
$$= \frac{(k+1)((k+1)+1)}{2} + (k+1)$$
therefor...
$$S_k+1 = Sk + (k+1)$$


Now I'm worried that I didnt really solve anything. I'm totally knew at this, and I'm open to criticism and help, just be kind :)
(ps.. this is my first time posting formulas, hopefully I did it right)


Thanks
dleacock

 

[dleacock]

just a repost to clean up the math...$$\frac{1}{4}(k+1)^2((k+1)+1)^2 + (k+1)^2$$

$$= \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2}$$

$$= \frac{(k+1)((k+1)+1)}{2} + (k+1)$$

 

[mattmns]

I don't own the book, but I looked at the pages on amazon, and it looks like you want to prove it using mathematical induction. This usually works in this way.

Proof:
Base Step (here, it will be n = 1)

Inductive Step Assume it is true for some k, and then show that it is true for k+1.
QED.Now you don't have to use induction, but it is what I think they want you to use. So first you should prove that the statement is true for n = 1, or k = 1. Meaning that if you plug 1 into the formula you will get the same thing, which is true.
$$1^3 = (1)^2$$
$$=> 1 = 1$$

So then do the inductive hypothesis.

You would say that the statement is true for some k, then show that it is true for k + 1.

This would mean that you would want to show that:

$$1^3 + 2^3 + 3^3 ... + k^3 + (k+1)^3 = (1 + 2 + 3 + ... + k + (k+1) )^2$$
Using the fact that the statement is true for k.

Meaning, prove the previous statement with the fact that: $$1^3 + 2^3 + 3^3 ... + k^3 = (1 + 2 + 3 + ... + k)^2$$

--------

just a repost to clean up the math...$$\frac{1}{4}(k+1)^2((k+1)+1)^2 + (k+1)^2$$

$$= \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2}$$

$$= \frac{(k+1)((k+1)+1)}{2} + (k+1)$$

I am not sure what you were doing here. But you can't just take the square root of something out of nowhere, also $$\sqrt{ a^2 + b^2 } \neq a + b$$ except in a few cases, so that is not allowed.

 

[dleacock]

yeah, that's exactly what I thouhgt the problem was. I just squared it out of nowhere. oh well, back to work I go. thanks for the response and the tip!

 

[ramsey2879]

You would say that the statement is true for some k, then show that it is true for k + 1.

This would mean that you would want to show that:

$$1^3 + 2^3 + 3^3 ... + k^3 + (k+1)^3 = (1 + 2 + 3 + ... + k + (k+1) )^2$$
Using the fact that the statement is true for k.

Meaning, prove the previous statement with the fact that: $$1^3 + 2^3 + 3^3 ... + k^3 = (1 + 2 + 3 + ... + k)^2$$

--------

In other words show, $$k^{2}(k+1)^{2} - k^{2}(k-1)^{2} = 4k^{3}$$