How to use a difference table to find a formula for a given sequence? Given a sequence, how do you get a "rule" from its difference table? For example;

1, 2, 3, 4, 5, 6
  1, 1, 1, 1, 1

Or

1, 2, 7, 16, 29
  1, 5, 9, 14
  4, 4, 4

I haven't really explored this topic before, and cannot find much online (I think I'm bad at searching)
Also I need this for a Lua program on iPod, if it's worth mentioning
 A: It’s really easy, if your sequence truly is given by a polynomial formula. First you have to know the Basic Functions $C_i$:
\begin{align}
C_0(n)&=1\quad\text{(constant)}\\
C_1(n)&=n\\
C_2(n)&=\frac{n(n-1)}2\\
C_3(n)&=\frac{n(n-1)(n-2)}{3!}\quad\text{etc.}
\end{align}
You see that these functions take integer values for all integer values of $n$, even though their coefficients are not integral.
Now you write your sequence and the sequence of successive differences. Rather than explain in the abstract what you do, I’ll use the sequence of where $F(n)$ is the sum of the first $n$ squares (starting at zero).
Our first sequence is $(0,1,5,14,30,\cdots)$, the sequence you’re interested in. The first differences are $(1,4,9,16,\cdots)$ (of course); the second differences are $(3,5,7,\cdots)$; and the third differences are constant $2$.
Now you look at the first entry in each list. These numbers are the coefficients of the functions $C_i$. In our example, these are $0$, $1$, $3$, and $2$. So your function is described as $C_1+3C_2+2C_3$. Expanding out you get
$$
F(n)=n + \frac{3n(n-1)}2+\frac{2n(n-1)(n-2)}6=\frac{2n^3+3n^2+n}6\,.
$$
That’s the method. Now you go home and give your own proof that it works.
A: By rule I think you meant the n$^{th}$ term of that sequence,
In any arithmetic sequence it is $\big(a+(n-1)d\big)$, where a is the first term and d is the common difference.
In your sequence it is, $\big(1+(n-1)1\big)=n$, that means that n$^{th}$ term of this sequence will be $n$.

In your second sequences the common difference between two consecutive term is in Arithmetic progression.
