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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.

  • 0
    @Ilmari: Good remark. I guess I was thinking about getting 5-$6$ decim$a$ls.2011-08-16

3 Answers 3

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$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 $

$= 110.485$.

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    +1 and accepted.This seems to be faster that my approach and also purports to be a general way,thanks :-)2011-08-16
4

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

$\begin{eqnarray*} S_{20} &=&\frac{1000}{9} -\frac{50}{81}+\frac{50}{81}10^{-20} \\ &=&\frac{220987654320987654321}{2000\,000\,000000\,000\,000}\approx 110.49. \end{eqnarray*}$

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    @FoolForMath: OK.2011-08-16
1

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} $

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    @Didier Piau: I guess I didn't notice where FoolForMath showed a method other than "after some algebra...". Since he was looking for something faster, I figured that whatever he did was not as fast. He also wanted to see why this was almost $110.5$, so I showed it was close to $110\frac{40}{81}$. If FoolForMath says this adds no value, I will delete my answer.2011-08-16