How to find a general sum formula for the series: 5+55+555+5555+.....? I have a question about finding the sum formula of n-th terms.
Here's the series:
$5+55+555+5555$+......
What is the general formula to find the sum of n-th terms?
My attempts:
I think I need to separate 5 from this series such that:
$5(1+11+111+1111+....)$
Then, I think I need to make the statement in the parentheses into a easier sum:
$5(1+(10+1)+(100+10+1)+(1000+100+10+1)+.....)$
= $5(1*n+10*(n-1)+100*(n-2)+1000*(n-3)+....)$
Until the last statement, I don't know how to go further. Is there any ideas to find the general solution from this series?
Thanks 
 A: 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...
A: 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.
A: Using the sum of a finite geometric series twice:
$$5+55+555+\ldots+\overbrace{55...5}^{n\;\text{times}}=\sum_{k=0}^n\left(5\cdot 10^k+5\cdot10^{k-1}+\ldots+5\cdot 10+5\right)=$$
$$=5\sum_{k=0}^n\frac{10^{k+1}-1}9=...\text{etc.}$$
A: $5+55+555+5555+55555+\cdots n$ terms,
Sum of finite seqence of this pattern can be obtained by re-arranging this into geometric sequence.
$5$ should be taken common first, we get $$5× (1+11+111+1111+11111+\cdots n)$$
$1, 11,111,\ldots$ are formed as quotients of chains of 9 and 9 like below
$$5× (9/9+99/9+999/9+9999/9+\cdots n)$$  then 9 is taken common out.
$$5/9×(9+99+999+9999+\cdots n)$$
$1,11,111,1111,...$are converted into constitution of geometric sequence as $$(10-1), (100-1), (1000-1)\ldots$$then
$5/9×(10-1+100-1+1000-1+10000-1+\cdots n)$ then its divided into two groups as below
$$5/9×(10+100+1000+10000+\cdots n -1-1-1-1-1-\cdots n)$$
first one is geometric sequence with common ratio 10 and first term akso 10 then using sum of finite GS for n terms $S_n = a(r*n - 1)/(r-1)$ for $r\neq1$ and for series $-1-1-1-1-1\cdots -n$ terms gets (-n) then using these we get,
$$5/9*(10*(10^n-1)/(10-1)-n)$$
$$5/9×(10/9×(10^n-1)-n)$$
$$50/81×(10^n-1)-5n/9$$ 
as required answer.. doing similarily we ca get sum of sequences like below
$0.4 + 0.44 + 0.444 + 0.4444 + \cdots$ upto n terms.
thank you ..
A: This might work: 

It's adding together powers of 10 multiplied by 5, and then adding together those numbers.
A: Obviously for the general term of the sum, $t_n=10t_{n-1}+5$, with $t_0=5$.
By identifying with $t_n+c=10(t_{n-1}+c)+5$, you can rewrite this recurrence as $$t_n+\frac59=10(t_{n-1}+\frac59),$$
then
$$t_n+\frac59=10(t_{n-1}+\frac59)=100(t_{n-2}+\frac59)...=10^{n}(t_0+\frac59)=10^n\frac{50}9.$$
Now you can sum these terms, using the formula for the geometric series:
$$\sum_{i=0}^{n-1} t_i=\sum_{i=0}^{n-1}\left(10^{n}\frac{50}9-\frac59\right)=\frac{10^n-1}{9}\frac{50}9-n\frac59.$$
A: $$\begin{align}5+55+555+5555+...&=5(1+11+111+1111+...)\\&=\frac 59(9+99+999+9999+...)\\&=\frac 59\sum_{i=1}^n 10^i-1\\&=\frac 59(\sum_{i=1}^n 10^i-\sum_{i=1}^n 1)\\&=\frac 59((\sum_{i=0}^n 10^i)-1-\sum_{i=1}^n 1)\\&=\frac 59\left(\frac {10^{n+1}-1}{10-1}-1-n\right)\\&=\frac 59\left(\frac {10^{n+1}-9n-10}9\right)\\&=\frac{50(10^{n}-1)-45n}{81}\end{align}$$
A: $$\begin{align}
S&=\;\;\;\underbrace{5+55+555+\cdots+\overbrace{555...55}^{n}}_{\text{$n$ terms}}\\
10S&=\quad\;\;\;\; \ 50+550+\cdots +555...50+\overbrace{5555...5}^{n}0\\
\\
10S-S&=-5\;\;-5\quad -5\quad\cdots\qquad\;-5+\overbrace{5555...5}^{n}0\\
\\
9S&=-5n+\overbrace{5555...5}^{n}0\\
\\
S&=\frac59 (\overbrace{1111...1}^{n}0-n)\qquad\blacksquare \end{align}$$
A: $$5+55+555+5555+\cdots+\overbrace{55\dots5}^{n\text{ fives}}$$
$$=\frac59(9+99+999+9999+\cdots+\overbrace{99\dots9}^{n\text{ nines}})$$
$$=\frac59(10^1-1+10^2-1+10^3-1+\cdots+10^n-1)$$
$$=\frac59(10^1+10^2+10^3+\cdots+10^n-n)$$
$$=\frac59\left(\frac{10^{n+1}-10}{9}-n\right).$$
A: $(\frac{5}{9})(10^n) - \frac{5}{9}$
$\frac{5}{9}$ is .555 repeating.  $10^n$ will shift the desired number of 5's past the decimal place.  Then subtracting $\frac{5}{9}$ will clip the remaining 5 repeating past the decimal place.
A: Here is a recurrence relation:


*

*$a_0=5$

*$a_n=10a_{n-1}+5(n+1)$


Converting this to a direct formula, you get: $$\frac{5\cdot10^{n+2}-45n-95}{81}$$
A: The first order difference of the series is $\Delta_n^1=55,555,5555,55555...$.
The second order difference is $\Delta_n^2=500,5000,50000...=50.10^n$.
As the finite difference of a power is the same power, we have (adding an affine term that vanishes with the second order difference)
$$S_n=a.10^n+bn+c,$$
$$\Delta_n^1=a.9.10^n+b,$$
$$\Delta_n^2=a.81.10^n.$$
Compute the unknowns from $\Delta_1^2=500, \Delta_1^1=55, S_1=5$ to get
$$S_n=\frac{50}{81}.10^n-\frac59n-\frac{50}{81}.$$
A: $$S = 5+55+555+5555+...$$
$$S = 5(1+11+111+1111+...)$$
$$S = 5\left(\sum_{k=0}^010^k+\sum_{k=0}^110^k+\sum_{k=0}^210^k+\sum_{k=0}^310^k+\dots \sum_{k=0}^n10^k\right)$$
Let
$$\sum_{k=0}^n10^k = \frac{1}{9}(10^{n+1}-1)$$
Which implies
\begin{align}
S &= 5\sum_{k=0}^n\left[ \frac{1}{9}(10^{k+1}-1)\right]\\
 &= \frac{5}{9}\sum_{k=0}^n(10^{k+1}-1)\\
 &= \frac{5}{9}\left(\sum_{k=0}^n10^{k+1}-\sum_{k=0}^n1\right)\\
 &= \frac{5}{9}\left(10\sum_{k=0}^{n}10^{k}-(n+1)\right)\\
 &= \frac{5}{9}\left(\frac{10}{9}(10^{n+1}-1)-n-1)\right)
\end{align}
Therefore

$$S = \frac{5}{81}\left(10^{n+2}-9n-19)\right)$$

