Solve the following recurrence relation: $S(1) = 2$; $S(n) = 2S(n-1)+n2^n, n \ge 2$ Solve the following recurrence relation:
$$\begin{align}
S(1) &= 2 \\
S(n) &= 2S(n-1) + n 2^n, n \ge 2
\end{align}$$

I tried expanding the relation, but could not figure out what the closed relation is:

+-------+------------------------------------------------+
|   n   |  S(n)                                          |
+-------+------------------------------------------------+
|   1   |  2                                             |
|   2   |  2S(1)+2*2^2 = 2*2 + 2*2^2 = 2^2 + 2*2^2 =     |
|       |              = 4 + 8 = 12                      |
|   3   |  2S(2)+3*2^3 = 2*(2^2 + 2*2^2) + 3*2^3 =       |
|       |              = 8 + 16 + 24 = 48                |
|   4   |  2S(3)+4*2^4 = 2*(2*2^2 + 2*2*2^2 + 3*2^3) +   |
|       |              + 4*2^4 =                         |
|       |              = 16 + 32 + 48 + 64 = 160         |
|   5   |  2S(4)+5*2^5 = 2*(2*2*2^2 + 2*2*2*2^2 +        |
|       |              + 2*3*2^3 + 4*2^4) + 5*2^5 =      |
|       |              = 32 + 64 + 96 + 128 + 160        |
|       |              = 384 + 96 = 480                  |
+-------+------------------------------------------------+


Edit: is it $S(n)=2^n\frac{n(n+1)}{2}$?


Given
$$
S(n)=2S(n-1)+n2^n\tag{1}
$$
Consider $T(n)=2^{-n}S(n)$. $T(1)=1$ and, multiplying $(1)$ by $2^{-n}$, we have the recurrence
$$
T(n)=T(n-1)+n\tag{2}
$$
Therefore, $T(n)=\frac{n^2+n}2$ and thus,
$$
S(n)=(n^2+n)\,2^{n-1}\tag{3}
$$
