Formula for calculating a progressive sum If we say that initially the addition is $1$, the sum $0$ and $d$ is constant of $5$.
step 1)
  sum = previous sum + addition            sum = 0 + 1 = 1
  addition = previous addition + d         addition = 1 + 5 = 6

step 2)
sum = previous sum + addition            sum = 1 + 6 = 7
addition = addition + d                  addition = 6 + 5 = 11

step 3)
sum = previous sum + addition            sum = 7 + 11 = 18
addition = addition + d                  addition = 11 + 5 = 16

step 4)... the same as above

What I want to achieve here is to find a formula for the calculations shown above from which I can find the sum if I know the rest.
The closest formula found so far is arithmetic progression. But still it's not the one I'm looking for.
What would be the formula for that?

First let's label the sums as a sequence $S_n$ and your "additions" as $A_n$.
It appears that $A_0$, the initial condition, is $1$, and $S_0=0$.
Then we have the system: $$S_{n+1} = S_n + A_n$$ $$A_{n+1} = A_n + D.$$
Here you have $D=5$. We can solve the recurrence for $A_{n+1}$ first, since it does not involve $S_n$.
Well, $A_0=1$, $A_1 = 1+D$, $A_2 = A_1 + D = 1 + 2D$, etc. Thus $A_{n} = 1 + nD$.
Plugging this back into $S_{n+1}$ gives:
$$S_{n+1} = S_n + 1 + nD.$$
Now $S_1 = 0 + 1 + D$, $$S_2 = S_1 + 1 + 2D = (0 + 1 + D) + 1 + 2D = 2 + (1+2)D.$$
$$S_3 = S_2 + 1 + 3D = (2 + (1+2)D) + 1 + 3D = 3 + (1+2+3)D$$ we can see the general form taking shape here:
$$S_{n} = n + (1+ 2+ 3+ \cdots + n)D= n + \frac{n(n+1)}{2} \cdot D.$$
