Q:

How to find the general formula of an arithmetic progression? 

I'm very new to combinatorics and graph theory and I was doing some work and I was asked to find certain values which I tabulated above. 
The table follows this general pattern and continues along. 
My problem is this;
How do I go about solving for the GENERAL FORMULA of tables of values such as this? 
OBSERVATION
I've made some deductions and with respect to the $D_1 \quad D_2 .....$ values etc  
So what I first did was collect ALL the values of $D_1$ in terms of $V$ and then $D_2$ in terms of $V$ etc until I reached $D_8$
The reason for focusing on the '$V$' value is that the graph theory work used to find the respective $D$ values is dependent upon $V$ , where $V$ denotes the vertex of the graph 
So for example $D_1 = V -1$
$D_2 = (V) + (V+1) + (V+2) + (V+3) + (V+4) + (V+5)....$
$D_3 = (V-1) + (V+1) + (V+3) + (V+5) +(V +7) + (V+9) ....$
And I did this all the way till $D_8$
Is this the correct method to try and find a formula to solve for the SB value ,assuming that this is a method , what do I need to do next? 
I need to find a formula such that when I enter the value of the vertex and edge I am able to compute the SB value in all cases. 
Can anyone point me in the right direction of what I need to do to achieve this? 
Is there any software that computes such a formula? 
What are some recommended reading materials to help me better solve these kinds of problems? 
The only book that I currently used is the Graph theory by Harary. 

A:

So your pattern seems to be that when $V = n+1$, $D_n = n$, and then it increases by $n$ each time.  In other words, when $V = n+k$, $D_n = kn$; or, in terms of $V$ directly, $D_n = (V-n)n = nV - n^2$.
The SB column is the sum $D_1 + D_2 + \dots + D_{V-1}$, giving us
$$
   \sum_{n=1}^{V-1} D_n = \sum_{n=1}^{V-1} (nV - n^2) = V \sum_{n=1}^{V-1} n - \sum_{n=1}^{V-1}n^2.  
$$
The first sum is the sum $1 + 2 + \dots + (V-1)$ and such sums are easy to compute: the values are uniformly spaced, and so we can take the average value (which is $\frac V2$) and multiply it by the number of terms (which is $V-1$).  We get $\frac{V(V-1)}{2}$ for the first sum, or $\frac{V^2(V-1)}{2}$ when you take the factor of $V$ in front into account. 
The second sum is the sum $1^2 + 2^2 + \dots + (V-1)^2$, and this is trickier to compute.  But there's a well-known formula for this sum: $$1^2 + 2^2 + \dots + (V-1)^2 = \frac{V(V-1)(2V-1)}{6}.$$ If you know the formula, then it can be verified using induction. 
Putting these together, we get
$$
   D_1 + D_2 + \dots + D_{V-1} = \frac{V^2(V-1)}{2} - \frac{V(V-1)(2V-1)}{6} = \frac{V(V-1)}{2} \cdot \left(V - \frac{2V-1}{3}\right) = \frac{(V+1)V(V-1)}{6}. 
$$
So that gives you the answer. 
However, since the formula we got is just a different way of writing $\binom{V+1}{3}$, you should see if there's a more direct argument for the thing you're counting: whatever it is, maybe you can interpret it as picking $3$ things out of $V+1$?