Closed form of $\sum_{k=0}^{n}k\cdot2^{n-k}$ I am interested in solving the recurrence relation using iterative method. I know how to solve it using generating function and another method using solution of associated linear homogeneous recurrence relation and particular solution described in Rosen's Discrete Math book. I came up with some summation (below) but I don't know how to find its closed form. I checked WolframAlpha and it provided me with this $$\sum_{k=0}^{n}k\cdot2^{n-k}=-n+2^{n+1}-2$$ How to show this?
HINT:

$$\sum_{k=0}^nk2^{n-k}=2^n\sum_{k=0}^nk\left(\frac12\right)^k\;,$$

so you’re done if you can evaluate

$$f(x)=\sum_{k=0}^nkx^k$$

at $x=\frac12$. Start with $g(x)=\sum_{k=0}^nx^k$, differentiate, and multiply by $x$. (I’m assuming that you know a closed form for $g(x)$.)
From $$ \sum_{k=0}^{n}x^{k}=\frac{1-x^{n+1}}{1-x}$$ 
differentiate to get$$ \sum_{k=0}^{n}kx^{k-1}=\frac{-nx^{n}\left(1-x\right)+1-x^{n}}{\left(1-x\right)^{2}}$$ 
Multiplying by $x$ so that $$ \sum_{k=0}^{n}kx^{k}=x\cdot\frac{-nx^{n}\left(1-x\right)+1-x^{n}}{\left(1-x\right)^{2}}$$ 
and replace $x$ by $\frac{1}{2}$, we get$$ \sum_{k=0}^{n}k\left(\frac{1}{2}\right)^{k}=-n2^{-n}+2-2^{-n+1}$$ Lastly, by multiplying $2^{n}$, $$ \sum_{k=0}^{n}k2^{n-k}=-n+2^{n+1}-2$$