Proving a complicated summation problem

I'm trying to solve a complicated summation problem where there are two multiplication problems that both have $$i$$. I'm entirely lost. I would be thankful for any help.

How is it possible to possible to find the sum of this expression (with a proof)?

$$\sum_{i=1}^{n} i(2^i)$$

\begin{align*} \sum_{i=1}^n i(2^i) &= 2^1 + 2\cdot2^2 + 3\cdot 2^3 + \cdots + n\cdot 2^n\\ &= \sum_{i=1}^n 2^i + \sum_{i=2}^n 2^i + \sum_{i=3}^n 2^i +\cdots + 2^n\\ &= (2^{n+1} - 2^1) + (2^{n+1}-2^2) + (2^{n+1} - 2^3) + \cdots + (2^{n+1} - 2^n)\\ &= n2^{n+1} - \sum_{i=1}^n 2^i = n2^{n+1} - (2^{n+1} - 2) = (n-1)2^{n+1} + 2 \end{align*}
Here is another approach, which yields a more general result. Start with the formula for the sum of a geometric series. $$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$$ Differentiate with respect to $$x$$. $$\sum_{i=0}^n i x^{i-1} = \frac{n x^{n+1}-(n+1)x^n+1}{(1-x)^2}$$ Multiply by $$x$$. $$\sum_{i=1}^n i x^i =\sum_{i=0}^n i x^i = x \left( \frac{n x^{n+1}-(n+1)x^n+1}{(1-x)^2} \right)$$ Now take the special case $$x=2$$. $$\sum_{i=1}^n i 2^i = 2\;[n 2^{n+1}-(n+1)2^n+1]$$