I am working on a summation of series.
Here is the Series
1- 10 0, 10 , 20, 30, 40 , 50, 60 ,70, 80, 90
11-20 100,120,140,160,180,200,220,240,260,280
21-30 300,340,380,420,460,500,540,580,620,660
As one can see, first diff is 10, second is 20 then 40 and next will be 80.
I want to perform a summation formula e.g if i place x = 54 i will get summation unto that point
I have tried doing stuff with n(n+1)/2 and similar summation formulae but can't get this thing in my mind yet
Let $A(n)$ be the $n$th number you add. The first thing to figure out is $A(10n+1)$ because things change there. Since the sum of powers of $2$ up to $2^n$ is $2^{n+1}-1$ we have $A(10n+1)=100(2^n-1)$
Then from $A(10n+1)$ to $A(10n+10)$ the increment is $10\cdot 2^n$.
The sum of elements from $A(10n+1)$ through $A(10n+10)$ is then $10A(10n+1)+45 \cdot 10 \cdot 2^n=1000(2^n-1)+450\cdot 2^n=1450\cdot 2^n-1000$.
The sum of all the elements up to $A(10n+10)$ is $1450(2^{n+1}-1)-1000(n+1)$
The sum of the elements from $A(10n+1)$ through $A(10n+m)$ is $100m(2^n-1)+\frac 12m(m-1)10\cdot 2^n=100m(2^n-1)+5m(m-1)\cdot 2^n$
So to get $A(k)$, express $k$ as $10n+m$ with $1 \le m \le 10$ and
$A(10n+m)=1450(2^{n}-1)-1000(n)+100m(2^n-1)+5m(m-1)\cdot 2^n$
