How to evaluate closed form of these series of sum? $\sum_{k=1}^n k*10^{k-1}$

Question

$$\sum_{k=1}^n k*10^{k-1}$$

I came across this summation of series while I was trying to solve Project Euler Problem 40. The problem can be solved without using this method; however, I want to know how to evaluate this summation to a formula. It gives 1, 21, 321, 4321, ... for n=1,2,3,4, ... $$\frac{10^n(9n-1)+1}{81}$$ I also obtained the formula by using symsum() function of Matlab, but I do not know how to evaluate it.

Answer 1

$$S(n)=\sum_{k=1}^n k10^{k-1}$$

$$=-(n+1)10^{n}+1+\sum_{k=1}^{n} (k+1)10^{k}$$

$$=-(n+1)10^n+1+\sum_{k=1}^{n} k10^k+\sum_{k=1}^{n} {10^k}$$

$$=-(n+1)10^n+1+10S(n)+\frac{10^{n+1}-1}{9}-1$$

Thus,

$$-9S(n)=-(n+1)10^n+1+\frac{10^{n+1}-1}{9}-1$$

$$S(n)=\frac{(n+1)10^n}{9}-\frac{10^{n+1}-1}{81}$$

$$=\frac{10^n(9n-1)+1}{81}$$

Answer 2

Hint. One may start with the standard geometric evaluation: $$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad x\ne1. \tag1 $$ Then by differentiating $(1)$ we have $$ 1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad x \ne1. \tag2 $$

Answer 3

$$\begin{matrix}1&+2\cdot10&+3\cdot100&+4\cdot1000&+\cdots&(n-1)\cdot10^{n-2}+&n\cdot10^{n-1}=\\\\ 1&+1\cdot10&+1\cdot100&+1\cdot1000&+\cdots&1\cdot10^{n-2}+&1\cdot10^{n-1}\\ &+1\cdot10&+2\cdot100&+3\cdot1000&+\cdots&(n-2)\cdot10^{n-2}+&(n-1)\cdot10^{n-1}\end{matrix}$$

So that, using the geometric summation formula for the first row,

$$S=\frac{10^n-1}{10-1}+10(S-n10^{n-1}).$$

Solving for $S$,

$$S=\frac{9n10^{n-1}-10^n+1}{81}.$$