How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$? 
Possible Duplicate:
Formula for calculating $\sum_{n=0}^{m}nr^n$ 

Can someone explain step by step how to derive the following identity?  $$\sum_{k=1}^{n} k \cdot 2^k = 2(n \cdot 2^n - 2^n + 1) $$

If you know $\sum_{k=1}^n a^k=\frac{a^{n+1}-a}{a-1}$, take the derivative of both sides with respect to $a$, multiply by $a$, and set $a=2$
