Given a sequence find nth term Suppose a sequence is given -: 3, 8, 20, 43, 83, 148 ...
We can find the differences between consecutive terms and do that over and over again until we have same differences (as Pascal's triangle)
3   8   20  43  83  148
 \ / \ / \ / \ / \ /
  5  12   23  40  65
   \ / \ / \ / \ /
    7   11  17  25
     \ / \ / \ /
      4   6   8
       \ / \ /
        2   2

How to find 'n-th term' and 'sum of first n term' of these kind of sequences?
 A: If we rotate the triangle $120^\circ$ (assuming the triangle is equilateral) clockwise, we have the following (original series shown in red; ignore items in blue and green for now):

$$\begin{array}
&&&&&&&&\\
&\color{blue}{a_6}&&a_5 &&a_4 &&a_3 &&a_2&&a_1&&a_0\\
&\color{blue}\downarrow&&\downarrow&&\downarrow &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\
&\color{blue} 0&&0 &&2 && 4 &&7 &&5 &&\color{red}3\\
&&\color{blue}0 &&2  &&6  &&11 &&12  &&\color{red}8 &\color{red}\uparrow\\
&&& \color{blue}2&&8  &&17 &&23  &&\color{red}{20}&\color{red}\uparrow&\color{red}{b_0}\\
&&&&\color{blue}{10}&&25  &&40 &&\color{red}{43}&\;\color{red}\uparrow&\color{red}{b_1}\\
&&&&&\color{blue}{35}&&65 &&\color{red}{83}&\;\color{red}\uparrow&\;\color{red}{b_2}\\
&&&&&&\color{blue}{100}&&\color{red}{148}&\;\color{red}\uparrow&\;\color{red}{b_3}\\
&&&&&&&\color{green}{248}&\;\color{red}\uparrow&\;\color{red}{b_4}\\
&&&&&&&\color{green}{\;\;\uparrow}&\;\color{red}{b_5}\\
&&&&&&&\color{green}{\;\;b_6}
\end{array}$$

In this triangle, any number is the sum of the two numbers above it and immediately to the left and right, similar to Pascal's triangle. Hence we have
$$
\begin{align}
b_0&=\binom 00a_0\\
b_1&=\binom 10 a_1+\binom 11 a_0\\
b_2&=\binom 20a_0+\binom 21a_1+\binom 22a_2\\
&\vdots \\
b_n&=\sum_{r=0}^n \binom nr a_r
\end{align}$$
Further, if the bottom row in the number triangle in the original question has elements of $2$ all the way, this means $a_r=0$ for $r\ge5$. We can then compute $b_n$ with only $a_0, a_1, a_2, a_3, a_4$, the values of which we already have.  
Hence, the general formula for $b_n$ becomes
$$\color{red}{b_n=\sum_{r=0}^{\min(4,n)} \binom nr a_r}\\
\color{red}{(a_r=3,5,7,4,2 \text{ for } r=0,1,2,3,4)}$$
This is also  general formula for $n$-th term of the original series (the first term indexed as $n=0$). 
Sum of first $n$ terms
The sum of the first $n$ terms is given by
$$\begin{align}
\sum_{j=0}^{n-1}b_j
&=\sum_{j=0}^{n-1}\sum_{r=0}^{\min (4,j)}\binom jr a_r\\
&=\sum_{r=0}^{\min(4,n-1)}  \sum_{j=r}^{n-1}\binom jr a_r\\
&=\color{red}{\sum_{r=0}^{\min(4,n-1)}  \binom n{r+1} a_r}\\
\end{align}$$
(see the solution here for an illustration)

Check 
As a quick check, we compute $b_6$ by hand (shown in blue and green in the diagram) and arrive at $248$.   
Using the formula, we have
$$b_6=\binom 61 \cdot 3 +\binom 62\cdot 5+\binom 63\cdot 7+\binom 64\cdot 4+\binom 65\cdot 2=248$$
as well.

Additional Note
It is interesting to note that we can compute $a_n$ directly from $b_r (r=0,1,\cdots n)$, using
$$a_n=\sum_{r=0}^n(-1)^{n+r}\binom nr b_r$$
For instance, $a_3=-1(3)+3(8)-3(20)+1(43)=4$.  
This is because the inverse of a Pascal matrix (a left-aligned Pascal's triangle) is also a Pascal matrix, but with elements of alternating sign!
(Discovered this in the course of solving this problem!)
A: Well, the bottom row is given by
$$a_n=2$$
So that the row above it is given by
$$b_n=2(n+1)$$
And the row above
$$c_n=5+n(n+1)$$
And
$$d_n=5n+\frac{n(n^2-1)}3$$
So finally, we have
$$e_n=3+\frac{5n(n-1)}2+\frac{n(n^2-1)(n-2)}{12}$$
And the sum of the first $n$ terms would be given by the row above that,
$$f_n=3n+\frac{5n(n-1)(n-2)}6+\frac{n(n^2-1)(n-2)(n-3)}{60}$$

As to how I found those, I used the fact that
$$x_k(n+1)-x_k(n)=x_{k-1}(n)$$
where
$$x_k(n)=\frac{(n+1)(n+0)(n-1)\dots(n+1-k)}{k!}$$
which is basically how your triangle was constructed.

Also, the sum of each line is equal to the above line plus a constant.
