Can You Solve This Exercise on Arithmetic-Geometric Series?

[Purpleshinyrock]

TL;DR

sequences,

Hello, I am currently self studying sequence and series and I got to a topic called arithmetic-geometric sequence, and after the theory It gives this exercise:

1) Find the sum:
S=1+11+111+1111+...+111...111, if the last (number) is a digit of n.

I was given a tip That says that
1 = (10 - 1)/9

11 = (100 - 1)/9 = (102 -1)/9

111 = (1000 - 1)/9 = (103 -1)/9

...

1111...111 = (100..000 - 1)/9 = (10n -1)/9

But I don't get how They got to this law of formation, did they apply a formula, what did they do?
Could You please help me?
Your time is appreciated.
Thank You.

 

[fresh_42]

Your sum is $1+11+111+\ldots+11\ldots 111 =\displaystyle{\sum_{k=1}^n \left(\dfrac{10^n}{9}-\dfrac{1}{9}\right)}$.

What do you know about $\sum_k (a_k+b_k)\, , \, \sum_k (c\cdot a_k)\, , \,\sum_k c$ and geometric series?

 

[Purpleshinyrock]

Your sum is $1+11+111+\ldots+11\ldots 111 =\displaystyle{\sum_{k=1}^n \left(\dfrac{10^n}{9}-\dfrac{1}{9}\right)}$.

What do you know about $\sum_k (a_k+b_k)\, , \, \sum_k (c\cdot a_k)\, , \,\sum_k c$ and geometric series?

I do not recognize the summation, And About the geometric series I know of their general term,common ratio, sum of elements

 

[fresh_42]

I do not recognize the summation, And About the geometric series I know of their general term,common ratio, sum of elements

$\sum_{k=1}^n (a_k+b_k)= (a_1+b_1)+(a_2+b_2)+\ldots +(a_n+b_n)$ explains the notation with $\Sigma$.
It is a short way to write sums without dots in between.

Given that, can you get a formula for:
\begin{align*}
\sum_{k=1}^n (a_k+b_k)& = \ldots \\
\sum_{k=1}^n (c\cdot a_k)& = \ldots \\
\sum_{k=1}^n c & = \ldots
\end{align*}
If you understand how sums of sums, sums of constant multiples, and sums of constants behave, then you can apply this to your formula. Finally you will need the summation formula for a geometric series:
$$
\sum_{k=1}^n q^n = \ldots
$$
These are the formulas you need to solve the question given the hint.