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
$$5+55+555+5555+\cdots+\overbrace{55\dots5}^{n\text{ fives}}$$
$$=\frac59(9+99+999+9999+\cdots+\overbrace{99\dots9}^{n\text{ nines}})$$
$$=\frac59(10^1-1+10^2-1+10^3-1+\cdots+10^n-1)$$
$$=\frac59(10^1+10^2+10^3+\cdots+10^n-n)$$
$$=\frac59\left(\frac{10^{n+1}-10}{9}-n\right).$$
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
$$S_n=\frac{50}{81}.10^n-\frac59n-\frac{50}{81}.$$