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?
 A: 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.
A: Let $S_n$ be sum to $n$ terms.
$$S_n = \frac 49 ((10-1) + (10^2-1) + ... + (10^n-1)) = \frac 49 \cdot (\frac{10\cdot(10^n-1)}{9}) - \frac{4n}{9}$$
Simplify.
A: We know that
$$ \sum_{k=0}^n {\frac{1}{x^k}} = 
   \frac{\frac{1}{x^{n+1}}-1}{\frac 1x - 1} = 
   \frac{x^{n+1}-1}{x^n(x-1)}$$.
Taking the derivative of both sides results in
$$\displaystyle \sum_{k=1}^n \frac{k}{x^{k+1}} =
  \frac{x^{n+1}-(n+1)x^n+n}{x^{n+1}(x - 1)^2}$$.
Multiplying both sides by $x^{n+1}$, we get
$$\sum_{k=1}^n kx^{n-k} = \dfrac{x^{n+1} - (n+1)x + n}{(x-1)^2}$$
Summing the first $n$ terms and letting $x = 10$ we get
\begin{align}
   4 + 44 + 444 + \dots + 444\ldots4
   &= 4(1 + 11 + 111 + \dots + 111\ldots1)\\
   &= 4[1+ (1+x) + (1+x+x^2) + \cdots + (1+x+x^2 + \cdots + x^{n-1})]\\
   &= 4[nx^0 + (n-1)x^1 + (n-2)x^2 + \cdots + 1x^{n-1}]\\
   &= 4[1x^{n-1}+ 2x^{n-2} + 3x^{n-3} + \cdots + nx^0]\\
   &= 4\sum_{k=1}^{n} kx^{n-k}\\
   &= 4\dfrac{x^{n+1} - (n+1)x + n}{(x-1)^2}\\
   &= \frac{4}{81}(10^{n+1} - 10(n+1) + n)
\end{align}
Checking the first few terms.
$4 = \frac{4}{81}(100 - 20 + 1) = 4$
$4 + 44 = \frac{4}{81}(1000 - 30 + 2) = 48$
$4 + 44 + 444 = \frac{4}{81}(10000 - 40 + 3) = 492$
