Using a number sequence as a row in Pascal's Triangle

Question

Background

Start with a sequence of real numbers: $x_0, x_1, x_2, ..., x_n$.

Treat this sequence as a row in Pascal's triangle and fill in the numbers below this row: that is, sum each consecutive pair of numbers in the sequence to create a new sequence, and repeat this process on each newly created sequence until only one number $y$ remains.

Example:

Sequence $= (5, 1, 3.2, -2)$

5     1     3.2   -2
   6    4.2    1.2
    10.2    5.4
       15.6

$y = 15.6$

Using the binomial theorem, I know that $y = \sum_{i=0}^{n} \binom{n}{i} x_i$ (and we can see that if every number in the sequence happens to be equal, then this becomes $y = 2^n * x_0$). Plugging in the above example, we get:

$ y = \sum_{i=0}^{n} \binom{n}{i} x_i = \binom{3}{0} * 5 + \binom{3}{1} * 1 + \binom{3}{2} * 3.2 + \binom{3}{3} * -2 = 1 * 5 + 3 * 1 + 3 * 3.2 + 1 * -2 = 15.6 $

I noticed that if the sequence is constructed such that the difference between any two consecutive numbers in the sequence is the same for all pairs of consecutive numbers (that is, there exists $c$ such that, for all $j \in \{1, 2, ..., n\}$, $x_j = x_{j - 1} + c$), that:

$y = \frac{2^n}{n+1}\sum_{i=0}^{n} x_i$, which I further simplified to $y = 2^n (x_0 + \frac{n}{2} c)$.

Example:

Sequence $= (6, 1, -4, -9)$

$c = -5$

$y = 2^n (x_0 + \frac{n}{2} c) = 2^3 (6 + \frac{3}{2} * -5) = 8 * -1.5 = -12$

6     1     -4    -9
   7     -3    -13
      4     -16
        -12

Note that if we have $c = 0$, then this becomes $y = x_0 * 2^n$, which as I noted above can be derived from the binomial theorem.

Question

How does this equation $y = 2^n (x_0 + \frac{n}{2} c)$ relate back to the binomial theorem? Why does it simplify so cleanly? Where does that $\frac{n}{2}$ come from?

Are there other ways to construct our sequence that also simplify in a similar way (for example, if we constructed it such that $x_j = c * x_{j-1}$, or $x_j = x_{j-1} ^ c$)?