Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS =  $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When 
$LHS_{P}$ = 
$RHS_{P}$
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(p+1)\cdot2^{p-1}=p\cdot2^{p}$
Case 3: n = p + 1
$LHS_{P+1}$ = $LHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = (p+1)$\cdot2^{p+1}$
So i need to to prove that:
$RHS_{P+1}$  = $RHS_{P}$  + (p+2)$\cdot2^{p}$
$RHS_{P+1}$  = $p\cdot2^{p}$  + (p+2)$\cdot2^{p}$
Am I thinking right here?
$RHS_{P+1}$ = 
$(p+1)2^{p+1}=$
$(p+1)\cdot2^{p}\cdot2$ = ? 
(Can I get this equal to  $p\cdot2^{p}$  + (p+2)$\cdot2^{p}$ ? )

Hint: You have to prove that $$n\cdot2^n +(n+2)\cdot2^n=(n+1)\cdot2^{n+1}$$
