# Talk:Sum of Sequence of Cubes/Proof using Multiplication Table

## Begging the question

In this section:

We have:
 $\ds 1^3 = 1$ $=$ $\ds \paren 2 \paren 1 \paren 1 - 1^2$

... etc.

isn't it being assumed what is trying to be proved? You want to show that $2 + 4 + 2 = 2^3$, $3 + 6 + 9 + 6 + 3 = 3^3$, etc. etc.

That is, you want to show that the outside edge is the cube.

So you can't do that by assuming it's $n^3$, what you need to do is show $n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n = n^3$

and you can't do that by assuming it equals $n^3$ on the left hand side.

So it would be best to remove those cubes from the LHS like it was when I had it.

The sum of cubes then falls out the bottom without problem.

It's the sum of all the squares equalling the square of the triangle number that needs a bit more rigour, as it's handwavey at the moment. --prime mover (talk) 16:26, 5 May 2020 (EDT)

Okay, so how do *you* explain the assumption that $n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n = n^3$ before you've proved it? You seem insistent on declaring they all equal cubes up front, I wonder how you "know beforehand" before it's proved. --prime mover (talk) 16:50, 6 May 2020 (EDT)

Got it. Thanks for the feedback.