Find the sum of $5.5+55.55+555.555..$ up till n terms? 
Find the sum of $5.5+55.55+555.555..$ up till n terms?

My attempt: $ 5.5+55.55+555.555 ... $
$ 5(1.1+11.11+111.111...) $
$ \dfrac{5}{9} \times 9(1.1+11.11+111.111..) $
$ \dfrac{5}{9} (9.9+99.99+999.999...) $
$ \dfrac{5}{9} (9+0.9+99+0.99+999+0.999...) $
$ \dfrac{5}{9} [(9+99+999 ...)+(0.9+0.99+0.999...)] $
$ \dfrac{5}{9} [(10-1+100-1+1000-1...)+(1-0.1+1-0.01+1-0.001...] $
So all the $1$ will cancel
$ \dfrac{5}{9} [(10+100+1000 ... n)+(n-(0.1+0.01+0.001...)] $
How to move forward? I can see two Geometric Progession in the 2 brackets but can't prove that? How do I continue? And is there any easier and less time taking method?
I have not as of yet learned summation.
 A: I just used the following summation:
\begin{align*}
\sum_{m=1}^n(5\sum_{i=1}^n(\frac{1}{10^i}+10^{i-1}))
\end{align*}
In my opinion it's a lot less hassle. A good rule of thumb is you pretty much always can find a summation for a sequence of added terms like this (and a product for multiplied ones!).
A: You've done a great job. The next thing is to continue the calculation. For instance, if n=5, then this would be 5/9(111110+5-1+0.88889)=5/9(111114.88889)=11111088889/180000
A: You have to make a general formula so you may use an already wxisting formula for sum of $n$ terms of  a $G.P.$
You arrived at $ \dfrac{5}{9} [(10+100+1000 ... n)+(n-(0.1+0.01+0.001...)] $
For a $G.P$ of type $a,ar,ar^2,\cdots ar^{n-1}$ The sum is given as
$S=a\displaystyle\frac{r^n-1}{r-1}$
In the first bracket $a=10$ and $r=10$ , for second bracket $ a=0.1$ and $r=0.1$ and hence we get
$\displaystyle S= \dfrac{5}{9} \bigg[(10*\frac{10^n-1}{10-1})+(n-\frac{1}{10}*\frac{1-\frac{1}{10})^n}{1-\frac{1}{10}})\bigg] $
$\displaystyle S= \dfrac{5}{9} \bigg[10*\frac{10^n-1}{9}+n-\frac{10^n-1}{9.10^n})\bigg] $
$S=\displaystyle\dfrac{50}{81}\bigg[10^n-1+\frac{9n}{10}-\frac{10^n-1}{10^{n-1}}\bigg]$
