Formula for calculating a progressive sum

Question

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

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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?

Answer 1

Let $a_n$ be the $n$-th sum and $b_n$ be the $n$-th addition.

From what you wrote, we have $$a_{n+1}=a_n+b_n\tag 1$$ $$b_{n+1}=b_n+5\tag 2$$ with $a_1=1,b_1=6$.

Now, from $(2)$, we have $$b_n=b_1+5(n-1)=5n+1.$$ Hence, from $(1)$, we have $$a_{n+1}-a_n=5n+1.$$ Hence, for $n\ge 2$, we have $$a_n=a_1+\sum_{k=1}^{n-1}(5k+1)=\frac{5n^2-3n}{2}.$$ This holds for $n=1$.

Answer 2

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:

$$S_{n} = n + (1+ 2+ 3+ \cdots + n)D= n + \frac{n(n+1)}{2} \cdot D.$$