A "fast" way to find the sum of the sequence $5,5.5,5.55,5.555,5.5555,\ldots $ (20 terms) My initial approach is diving the whole sum by $9$ and taking the common $5$ out which gives $$\frac{5}{9}[(10-1)+(10-0.1)+(10-0.01)+\cdots + (10-10^{-19})]$$ after some algebra this could be reduced to $$\frac{5}{9} \times [200-\frac{10}{9} \times (1-10^{-20})]$$
after this I am not sure how to show that is almost equal to $110.5$? Also if any body wants to suggest any other tricky/fast way I will appreciate it.
 A: $5+5.5+5.55+5.555+\cdots $
$= 5 + (5+0.5) + (5+0.5+0.05) + \cdots$
$= 20 \times 5 + 19 \times 0.5 + 18 \times 0.05 + 17 \times 0.005 \cdots + 1 \times 0.00{\ldots}05$
$\approx 100+9.5+0.9+0.085 $
$= 110.485$.
A: A slower way. It solves the general case with $n$ terms. I denote your sum as
$$S_{20}=5+5.5+5.55+5.555+\ldots +5.\underset{19}{\underbrace{555\ldots 5}}$$
and the general case for $n$ as
$$S_{n}=5+5.5+5.55+5.555+\ldots +5.\underset{n-1}{\underbrace{555\ldots 5}}.$$
Since $\sum_{j=1}^{k-1}10^{j-1}=\frac{1}{90}10^{k}-\frac{1}{9}$, we have in general
$$\begin{eqnarray*}
S_{n} &=&5n+5\sum_{k=1}^{n}\frac{\sum_{j=1}^{k-1}10^{j-1}}{10^{k-1}} =5n+5\sum_{k=1}^{n}\frac{\frac{1}{90}10^{k}-\frac{1}{9}}{10^{k-1}} 
=\frac{1000}{9}-\frac{50}{81}+\frac{50}{81}10^{-n},
\end{eqnarray*}$$
and in the present case
$$\begin{eqnarray*}
S_{20} &=&\frac{1000}{9} -\frac{50}{81}+\frac{50}{81}10^{-20} \\
&=&\frac{220987654320987654321}{2000\,000\,000000\,000\,000}\approx 110.49.
\end{eqnarray*}$$
A: As an alternative you could try
$$
\begin{align}
\sum_{k=0}^{19}(\frac{50}{9}-\frac{5}{9}10^{-k})&=\frac{50}{9}\times 20-\frac{5}{9}\frac{1-10^{-20}}{1-10^{-1}}\\
&=\frac{1000}{9}-\frac{50}{81}\left(1-10^{-20}\right)\\
&=\frac{8950}{81}+\frac{50}{81}\times 10^{-20}\\
&=110\frac{40}{81}+\frac{50}{81}\times 10^{-20}
\end{align}
$$
