Answer
See below.
Work Step by Step
1. For $n=0$, we have $LHS=1(2)^1=2$ and $RHS=0+2=2=LHS$, thus P(0) is true.
2. Assume P(k) ($k\gt0$) is true, we have:
$\sum_{i=1}^{k+1}i\cdot 2^i=k\cdot 2^{k+2}+2$
3. For $n=k+1$, we have:
$LHS=\sum_{i=1}^{k+1}i\cdot 2^i+(k+2)2^{k+2}=k\cdot 2^{k+2}+2+(k+2)2^{k+2}\\
=2^{k+2}(2k+2)+2=(k+1)2^{k+3}+2=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.