Re: Differences of differences solution.

In the last post, the one just below this one, 

I wrote about how frustratingly difficult it is 

to solve the differences of differences problem. 

Although this isn't a complete proof, I decide to 

go brute force and solve 'em one-by-one. I am 

starting with a line of squares, so here goes:

1   4   9   16   25...

  3   5    7    9...

     2   2   2...

Now although I got a line of 2s for the first 5 square 

numbers, I have to show it applies for the rest of the

sequence. I do this algebraically:

Let the first number be n².
The second number becomes(n+1)².
Difference= 2n+1.

'First number' can be generalised to 'any 
number in the sequence', and since 2n+1 
represents an odd number, increasing ns 
will produce a line of consecutive odd 
numbers. And what is the difference between 
consecutive odd numbers? Exactly, 2.

So that proves the case for 2, but it is nowhere near a complete proof.
It might be possible to do that with cubes:

Let the first number be n^3.
The second number becomes(n+1)^3.
Difference= 3n²+3n+1.

That's a line of (3n²+3n+1), with an 
increasing n.
New sequence:

Let the first number be 3n²+3n+1.
The second number becomes 3(n+1)²+3(n+1)+1.
Difference= 6n+6.
Now with 6n+6, the difference = 6.
So that solves it for 3.

I'm guessing you can also do that for higher bases, but you can't 
keep doing it for every exponent.