Find the sum of the n first series numbers: $7,77, 777,...$ Find the sum of the $n$ first numbers: $7,77, 777,...$
I thought to find an order by dividing $77/7=11, 777/7=111...$
but I don't know how to continue.
 A: JimmyK4542 gave a great answer. I will break it down a little bit more. Write 
$$\begin{align*}
\underbrace{7+77+777+\ldots}_{n\text{ terms}} &= \frac{7}{9}\left(\underbrace{9+99+999+\ldots}_{n\text{ terms}}\right) \\\\
&= \frac{7}{9}\Big(\underbrace{(9+1)+(99+1)+(999+1)+\ldots}_{n\text{ terms}}-n\Big) \\\\
&= \frac{7}{9}\left(\sum_{k=1}^n10^k-n\right)\;.
\end{align*}$$ 
And lastly use the fact that $$\sum_{k=1}^nx^k = \frac{x^{n+1}-1}{x-1}-1$$ with $x = 10$ to give yourself a closed form solution.
A: Hint: If you divide each term by $7$, then multiply each term by $9$, the series becomes: $9, 99, 999, 9999, \ldots$. 
Now, add $1$ to each term to get a familiar looking series: $10, 100, 1000, 10000, \ldots$. 
The $n$-th term of this series is clearly $10^n$. Now, undo those operations to get an expression for the $n$-th term of the original series. Finding the sum of this series should be easy if you know the formula for the sum of an geometric series.
A: Hint:
Write $u_n = 7+\cdots+777$ ($n$ terms). You know $u_0,u_1$; moreover, $u_{n+1} = 10u_n + 7(n+1)$ (can you see why?).
A: In this series , $$7_{⠀⠀⠀70⠀⠀⠀} 77_{⠀⠀⠀700⠀⠀⠀} 777_{⠀⠀⠀7000⠀⠀⠀}.⠀⠀.⠀⠀⠀.⠀⠀⠀T_n$$
So , here  $T_n$ will be first term of the series plus sum of the differences of consecutive terms . We also know that : $$T_{n}-T_{n-1}=7\cdot10^{n-1}$$ We can say that the difference is in $GP$ . $$T_n=7+\frac{70(10^{n-1}-1)}{9}$$
Hence ,
$$S_n=\sum_{r=1}^{n}T_r\Rightarrow \sum_{r=1}^{n}\left(7+\frac{70(10^{r-1}-1)}{9}\right)=7n+\frac{70}{9} \sum_{r=1}^{n}(10^{r-1}-1) \\
\Rightarrow 7n+\frac{70(1+10^1+10^2+.......10^{n-1}-n)}{9} \Rightarrow 7n+\frac{70}{9}\left(\frac{10^n-1}{9}-n\right)$$
So , Finally

$$S_n=7n+\frac{70}{9}\left(\frac{10^n-1}{9}-n\right)$$

