TITLE: Finding the nth term of a numeric sequence- Newton's little formula explanation
QUESTION [5 upvotes]: In a contest problem book, I found a reference to Newton's little formula that may be used to find the nth term of a numeric sequence. Specifically, it is a formula that is based on the differences between consecutive terms that is computed at each level until the differences match. 
An example application of this formula for computing the nth term of the series (15, 55, 123, 225, 367, 555, 795, ....) involves computing the differences as shown below:  

1) 1st Level difference is (40, 68, 102, 142, 188, 240)
2) 2nd Level difference is (28, 34, 40, 46, 52)
3) 3rd Level difference is (6, 6, 6, 6, 6) 

Now the nth term is  $$15{n-1\choose 0} + 40{n-1\choose 1} + 28{n-1\choose 2} + 6{n-1\choose 3}$$ where the constant multipliers are the first term of the differences at each level in addition to the first term of the sequence itself.
I was not able to find any reference to this formula or a proof of it after searching on the web. Any explanation of this method is appreciated.

REPLY [4 votes]: 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.