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

Question

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.

Answer 1

We are to find the value of $5.5 + 55.55 + 555.555 + \cdots$ from the first term until the $n$th term.

Now, as you did, it can be factored into $5(1.1 + 11.11 + 111.111 + \cdots)$. Ignoring the decimal point for now, we can see that the terms are $11, 1111, 111111, \cdots$ where the $i$th term is $$1 + 10 + 10^2 + \cdots + 10^{2i}.$$ This is clearly a geometric progression with initial value $1$ and common ratio $10$. The number of terms in this series is $2i$. Now, using the formula to get the sum of a geometric progression,

\begin{align*} S &= a_1\left(\frac{1 - r^n}{1 - r}\right) \\ &= \frac{1 - 10^{2i}}{1 - 10} \\ &= \frac{10^{2i} - 1}{10 - 1} \\ &= \frac{1}{9}(10^{2i} - 1) \end{align*}

Now, the number of times to divide $10$ seems to be $i$. Therefore, the $i$th term in the series $1.1 + 11.11 + 111.111 + \cdots$ is $$\frac{1}{9(10^i)}(10^{2i} - 1).$$ We can rewrite this as $$\frac{1}{9}\left(10^{i} - \frac{1}{10^{i}}\right).$$ Now, solving for the value of the series, we can separate terms and get $$\frac{1}{9}\left(10 + 10^2 + 10^3 + \cdots + 10^n\right) - \frac{1}{9}\left(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \cdots \frac{1}{10^n}\right) \\ \frac{1}{9}\left(10\left(\frac{10^n - 1}{10 - 1}\right)\right) - \frac{1}{9}\left(\frac{1}{10}\left(\frac{(\frac{1}{10})^n - 1}{\frac{1}{10} - 1}\right)\right) \\ \frac{10}{81}(10^n - 1) - \frac{1}{810}\left(\frac{10^n - 1}{10^{n - 1}}\right) \\ \frac{10^n - 1}{81}\left(10 - \frac{1}{10}\left(\frac{1}{10^{n - 1}}\right)\right) \\ \frac{10^n - 1}{81}\left(10 - \frac{1}{10^{n}}\right) \\ \frac{1}{81}\left(\frac{(10^n - 1)(10^{n + 1} - 1)}{10^n}\right)$$ Lastly, we multiply this value by $5$ and we get $$\frac{5}{81}\left(\frac{(10^n - 1)(10^{n + 1} - 1)}{10^n}\right).$$

Answer 2

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!).

Answer 3

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

Answer 4

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] $

$S=\displaystyle\dfrac{50}{81}\bigg[10^n-1+\frac{9n}{10}-\frac{10^n-1}{10^{n-1}}\bigg]$