How to find a general sum formula for the series: 5+55+555+5555+.....?
Solution 1:
$$5+55+555+5555+\cdots+\overbrace{55\dots5}^{n\text{ fives}}$$ $$=\frac59(9+99+999+9999+\cdots+\overbrace{99\dots9}^{n\text{ nines}})$$ $$=\frac59(10^1-1+10^2-1+10^3-1+\cdots+10^n-1)$$ $$=\frac59(10^1+10^2+10^3+\cdots+10^n-n)$$ $$=\frac59\left(\frac{10^{n+1}-10}{9}-n\right).$$
Solution 2:
Using the sum of a finite geometric series twice:
$$5+55+555+\ldots+\overbrace{55...5}^{n\;\text{times}}=\sum_{k=0}^n\left(5\cdot 10^k+5\cdot10^{k-1}+\ldots+5\cdot 10+5\right)=$$
$$=5\sum_{k=0}^n\frac{10^{k+1}-1}9=...\text{etc.}$$
Solution 3:
Here is a recurrence relation:
- $a_0=5$
- $a_n=10a_{n-1}+5(n+1)$
Converting this to a direct formula, you get: $$\frac{5\cdot10^{n+2}-45n-95}{81}$$
Solution 4:
Hint: Rather than gathering the terms in $10^r$ together as you have, try first summing $10^{n-1}+10^{n-2}+\dots +1$ as a geometric progression - which should give you a term in $10^n$ plus a constant. The constants are easy to add, and the terms in $10^n$ are another geometric progression.
Solution 5:
I think something like
$$5\sum_{i=0}^n (n+1-i)10^{i}$$
should work.
Some explanation as how it works: First of all I rewrite as: $5(1 + 11 + 111 + \ldots)$. Then notice that I can construct this sum also with adding numbers of the form $10^i$, while considering that at each digit the number should get added multiple times depending on the length of the number. For example:
- $n=0$: $\,5 \cdot 10^0 = 5$
- $n=1$: $\,5 \cdot (2 \cdot 10^0 + 1 \cdot 10^1)= 5 \cdot (2 + 10) = 60$
- $n=2$: $\,5 \cdot (3 \cdot 10^0 + 2 \cdot 10^1 + 1 \cdot 10^2)= 5 \cdot (3 + 20 + 100) = 615$
and so on...