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?
Best Answer
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 $n=1$.