Q:

Proof for : How many numbers cannot be expressed as a sum of two numbers which are chosen from a given set.

We are given two numbers $a,b$ and  a range of numbers from $[b,a+b-1]$ where $b\geq1$ and $a\gt2 $. We can add any two numbers $p,q$ (not necessarily distinct) in the range to form more numbers and add them to the set. How many numbers cannot be formed even after performing these operations infinite number of times.
My Approach is this:
Two cases arise when $a\geq b$  and  $a\lt b$
1.) $a\geq b$ , here the answer is simply $b-1$.
2.)$a\lt b$ is where I have found a series but it may always not satisfy every case.  The series is an Arithmetic progression with last term as $ b-1 $ , common difference as $ a-1 $ and number of terms  as $b-1/a-1$ .
 I am not sure about this formula and hence would appreciate a mathematical proof.
3.) There is also a special case when $a=2$ , then for some $b$ the number of values which are cannot be formed gives us the the triangular number series i.e $(1,3,6,10,...)$
but this would be handled by a generalised formula , which i am still doubtful about. 
Any proofs or suggestions ? 

A:

Initially, you have the set of $S_1 = [b, b + (a-1)]$. During the first set of additions, the range of new values will be from the smallest value added to itself to the largest value added to itself, i.e., $S_2 = [2b, 2b + 2(a-1)]$. The second set of additions generate the additional sets of $S_3 = [3b, 3b + 3(a-1)]$ when adding between $S_1$ and $S_2$, plus $S_4 = [4b, 4b + 4(a-1)]$ when adding among $S_2$. Next, with the third set of additions, you can get $S_5 = [5b, 5b + 5(a-1)]$ by adding among $S_1$ and $S_4$ or $S_2$ and $S_3$, and $S_6 = [6b, 6b + 6(a-1)]$ by adding among $S_3$. Continuing these additions, the final set $S$ would be the union of all these sub-sets $S_i$.
You can use Complete (strong) induction to prove each new sub-set $S_k$ of values of the form $kb$ plus a non-negative integer is $[kb, kb + k(a-1)]$. This has already been shown for $k = 1$. Assume it's true for all $k \le n$ for some $n \ge 1$. For $S_{n+1}$, note the values to get $(n+1)b$ come from $S_j$ and $S_m$ where $(j + m)b = (n + 1)b$, i.e., $j + m = n + 1$. The bottom values are $jb$ and $mb$, so their sum is $(j+m)b = (n+1)b$, and the top values for each is $jb + j(a - 1)$ and $mb + m(a - 1)$, so their sum is $(j+m)b + (j + m)(a - 1) = (n+1)b + (n + 1)(a - 1)$. Thus, the new sub-set is $S_{n+1} = [(n+1)b, (n+1)b + (n+1)(a-1)]$. This proves the inductive step.
As for how many positive integers are not included, they will be from just after the top of each range to just before the start of the next range until the ranges overlap. In particular, for range $i$, the number of values is $((i + 1)b - 1) - (ib + i(a - 1) + 1) + 1 = b - i(a - 1) - 1$. Note this also works for before $b$ as $i = 0$ gives $b - 1$. Thus, in general, the total number $N$ is
\begin{align}
N & = \sum_{i=0}^k (b - i(a - 1) - 1) \\
& = (k+1)(b-1) - \left(\frac{k(k+1)}{2}\right)(a-1) \\
& = (k+1)\left(b - 1 - \frac{k(a-1)}{2}\right) \tag{1}\label{eq1}
\end{align}
where $k$ is the largest integer where $b - k(a - 1) - 1 > 0$. This can be manipulated to show that $k$ is the largest integer where $k \lt \frac{b-1}{a-1}$. It can be expressed more succinctly by
$$k = \left\lfloor \frac{b-2}{a-1} \right\rfloor \tag{2}\label{eq2}$$
For your case $(1)$ where $a \ge b$, then $b - 2 \lt a - 1$ so \eqref{eq2} gives $k = 0$, and thus \eqref{eq1} gives $N = b - 1$ as you stated.

