Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + n \cdot 2^{n} = (n-1) 2^{n+1} + 2$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + n \cdot 2^{n} = (n-1) 2^{n+1} + 2$
For $n=1$,we have
$P(1): 1 \cdot 2 = 2 = (1-1) 2^{1+1} + 2 = 0 + 2 = 2$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + k \cdot 2^{k} = (k-1) 2^{k+1} + 2$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$\{1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + k \cdot 2^{k}\} + (k+1) \cdot 2^{k+1}$
$= (k-1) 2^{k+1} + 2 + (k+1) 2^{k+1}$
$= 2^{k+1} \{(k-1) + (k+1)\} + 2$
$= 2^{k+1} \cdot (2k) + 2$
$= k \cdot 2^{(k+1)+1} + 2$
$= \{(k+1)-1\} 2^{(k+1)+1} + 2$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.