Sum the series

The Sum of 1 + 22 + 333 + 4444 + ... to $n$ terms

This is an extension of the NRICH problem called Clickety Click and All the Sixes , where the solution involves summing the series

$$1+11+111+1111+\cdots$$ to $n$ terms.

Each term of the series $1 + 22 + 333 + 4444 +\cdots$ can be written as $k (1 + 10 + 100 + \cdots + 10^{k -1} )$ for some value of $k$. Using geometric series theory each term can be written as $k (10^ {k} -1)/9$. So the sum $S_n$ can be written as $$\begin{array}{rll}{S}_n& =& 1\left(\frac{{10}^1-1}{9}\right)+2\left(\frac{{10}^2-1}{9}\right)+3\left(\frac{{10}^3-1}{9}\right)+4\left(\frac{{10}^4-1}{9}\right)+\cdots \\ & =& \frac{1}{9}({1.10}^1+{2.10}^2+{3.10}^3+{4.10}^4+\cdots +n{.10}^n-(1+2+3+4+\cdots +n))\\ & =& \frac{1}{9}({1.10}^1+{2.10}^2+{3.10}^3+{4.10}^4+\cdots +n{.10}^n-\frac{1}{2}n(n+1))\end{array}$$

To find the formula for $1.10^1 + 2.10^2 + 3.10^3 + 4.10^4 + \cdots + nx^n$ consider the series $$y=1+x+x^2+x^3+\cdots +x^n=\frac{x^{n+1}-1}{x-1}$$

Differentiating this expression and multiplying the derivative by x we get $$\begin{array}{rll}x\frac{dy}{dx}& =& x+2x^2+3x^3+\cdots +n{x}^n\\ & =& \frac{n{x}^{n+2}-(n+1)x^{n+1}+x}{(x-1{)}^2}\end{array}$$.

Hence $$10+{2.10}^2+{3.10}^3+\cdots +n{.10}^n=\frac{n{.10}^{n+2}-(n+1){.10}^{n+1}+10}{81}$$

So the sum of $n$ terms of the original series is $$S_n=1+22+333+4444+\cdots =\frac{1}{9}(\frac{n{.10}^{n+2}-(n+1){.10}^{n+1}+10}{81}-\frac{1}{2}n(n+1))$$

Note that when $n$ is greater than 9, terms cannot be written with a repeated single digit and the $k^{th}$ term should be treated as $k (10^0 + 10^1 + 10^2 + \cdots + 10^{k-1})$.