Finding the nth term of a numeric sequence- Newton's little formula explanation
Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:
$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$
Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:
$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$
Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.
Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$
If it is the case you want the sequence to start with $a_1$ we need to shift this result to the right one:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(1) {n-1 \choose i}$$
Note when you re-define $a_0$ to be $a_1$ by shifting it's index to the right one, you're re-defining $a_1-a_0=\Delta^1(0)$ to be $a_2-a_1=a_{1+1}-a_{1}=\Delta^1(1)$.
Starting with the second differences we have
- $28 = 28$,
- $34 = 28 + 6$,
- $40 = 28 + 6 + 6$
- $46 = 28 + 6 + 6 + 6$$
Now for the first differences we have
- $40 = 40$
- $68 = 40 + 28$
- $102 = 40 + 28 + 34 = 40 + 28 + (28 + 6)$
- $142 = 40 + 28 + 34 + 50 = 40 + 28 + (28 + 6) + (28 + 6 + 6)$
So the original sequence is
- $15 = 15$
- $55 = 15 + 40$
- $123 = 15 + 40 + 68 = 15 + 40 + [40 + 28]$
- $225 = 15 + 40 + 68 + 102 = 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)]$
- \begin{align} 367 &= 15 + 40 + 68 + 102 + 142 \\ &= 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)] + [40 + 28 + (28 + 6) + (28 + 6 + 6)] \end{align}
The patterns should be clear. So just count how many 15's, 40's, 28's, and 6's there are in each term.