Seeking a combinatorial proof of the identity $1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$ I would appreciate if somebody could help me with the following problem
Q: show that combinatoric identity (using by combinatorial proof)
$$1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$$

Combinatoric interpretation.
Sub-task:

There is a ordered sequence (array, vector, chain, cortege, ...) of $k$ balls.
1 ball $-$ red, other balls $-$ black or white.
Let $M(k)$ $-$ number of such arrays.  
$M(k) = k \cdot 2^{k-1}$.
Example for $k=3$:
$\Huge{\color{red}\bullet} \circ\circ$,
$\Huge{\color{red}\bullet} \circ\bullet$,
$\Huge{\color{red}\bullet} \bullet\circ$,
$\Huge{\color{red}\bullet} \bullet\bullet$;

$\Huge{\circ\color{red}\bullet} \circ$,
$\Huge{\circ\color{red}\bullet} \bullet$,
$\Huge{\bullet\color{red}\bullet} \circ$,
$\Huge{\bullet\color{red}\bullet} \bullet$;
$\Huge{\circ\circ\color{red}\bullet}$,
$\Huge{\circ\bullet\color{red}\bullet}$,
$\Huge{\bullet\circ\color{red}\bullet}$,
$\Huge{\bullet\bullet\color{red}\bullet}$.
Task:

To find $S(n)$: number of all possible arrays (with 1 red and other black/white balls)  with length $k\leqslant n$.
$S(n) = M(1)+M(2)+\ldots+M(n)$.

How to prove:
Let $L(k)$ $-$ number of arrays, where red ball is on the $k$-th place:

$L(1) = 1+2+4+\cdots+2^{n-1} = 2^n-1$;
$L(2) = 2+4+\cdots+2^{n-1} = 2^n-2^1$;
$\ldots$
$L(k) = 2^{k-1}+2^k+\cdots+2^{n-1} = 2^n-2^{k-1}$;
$\ldots$
$L(n-1) = 2^{n-2}+2^{n-1} = 2^n-2^{n-2}$;
$L(n) = 2^{n-1} = 2^n-2^{n-1}$.
So, 
$$
S(n) = \sum_{k=1}^n L(k) = \sum_{k=1}^{n} (2^n - 2^{k-1}) = n2^n - (2^n-1) = (n-1)2^n+1.
$$
