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?  $$\sum_{k=1}^{n} k \cdot 2^k = 2(n \cdot 2^n - 2^n + 1) $$
 A: 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 $a=2$
A: Although you got great answers, I will add another way to look at the expression you are looking for
Let us denote
$$ S = \sum_{k=1}^{n} k \cdot 2^k $$
This can be expanded as 
$$ 
\begin{align*}
  S &= 2 + 2 . 2^2 + 3 . 2^3 + \cdots n 2^n\\
2S &= \hspace{8pt}+1.2^2+2.2^3 + \cdots (n-1) . 2^n+n . 2^{n+1}
\end{align*}
$$ 
If you notice, by multiplying by $2$ and writing under similar terms and then subtracting, we get
$$
\begin{align*}
S = n . 2^{n+1} - \left(2+2^2+2^3+\cdots+2^n \right) &= n .2^{n+1}-2^{n+1}+2\\
&= 2(n . 2^n-2^n+1)
\end{align*}
$$
A: Induction is a surefire way to prove it with elementary methods. With calculus and some formulas, we can approach this problem by evaluating a derivative in two different ways, as follows:
Quotient rule: $$\rm \frac{d}{dx}\frac{x^n-1}{x-1} =\frac{n\,x^{n-1}(x-1)-(x^n-1)(1)}{(x-1)^2} \tag{1}$$
Geometric formula:
$$\rm \frac{d}{dx}\frac{x^n-1}{x-1}=0+1x^0+2x^1+\cdots+n\,x^{n-1} \tag{2}$$
Equate equation $(1)$ with equation $(2)$, multiply both sides by $\rm x$ and then set $\rm x=2$.
