Find the partial sums of $4+44+444+\cdots$ 
Find the sum to $n$ terms of $4+44+444+\cdots $

My attempts:


*

*Used successive difference method

*Used $4+(4+40)+(4+40+400)\dots$ method
Failed to get a formula for partial sum in both ways. What am I missing?

We can Write $\displaystyle 4 = 4(1) = \frac{4}{9}\left[9\right] = \frac{4}{9}\left[10-1\right]$
Similarly $\displaystyle 44 = 4(11)=\frac{4}{9}\left[99\right] = \frac{4}{9}\left[10^2-1\right]$
Similarly $\displaystyle 444 = 4(111)=\frac{4}{9}\left[999\right] = \frac{4}{9}\left[10^3-1\right]$
Now you can proceed from here.
