Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ 
Possible Duplicate:
How to compute the  formula $\sum  \limits_{r=1}^d r \cdot 2^r$? 

How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, but by result $2^n(n+2)$ isn't correct according to Wolfram ($2^n (n-2)+2$). 

$$x+x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x}{x-1}$$
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$$0x+x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x^2}{x-1}$$
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$$0x+0x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x^3}{x-1}$$
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$$0x+0x^2+0x^3+...+0x^{n-1}+x^n=\frac {x^{n+1}-x^n}{x-1}$$
After adding we get:
$$x+2x^2+...+nx^n=\sum_{i=1}^{n}\frac {x^{n+1}-x^i}{x-1}=\frac{nx^{n+1}}{x-1}-\sum_{i=1}^{n}\frac {x^i}{x-1}=\frac{nx^{n+1}}{x-1}-\frac {x^{n+1}-x}{(x-1)^2}$$
