I have this problem. If $y=2x^3-x^2+3x+1$ Calculate the values of y corresponding to $x=0,1,2,3,4,5$ and form the table of forward differences. Prove theoretically that the second differences $12x+10$ and verify this numerically.
(So the issue is not making the forward differences table it's the part with proving. How should I go about this? I can post a picture of the table if needed.
You can find the first differences by simplifying the following, which is the general formula for the $x^{th}$ first difference:
$2(x+1)^3 - (x+1)^2 + 3(x+1)+1 - (2x^3 - x^2 + 3x+1)$
Basically, get the $(x+1)^{th}$ value of y and subtract the $x^{th}$ value of y.
This gives you $6x^2+4x+4$
Repeating this process to find the $x^{th}$ second differences, meaning simplifying the following
$6(x+1)^2+4(x+1)+4-(6x^2+4x+4)$
gives you the solution. (Note: I'd prefer to use n instead of x, but you used x in your question so I stuck with that.)