Induction Proof: $\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $ Prove by Mathematical Induction . . .
$$\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $$
for all $n \geq 0$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:

$1\cdot 2^1 = 0\cdot 2^{0+2}+2$
$2 = 2$

b. Inductive Hypothesis

$$\sum_{i=1}^{k+1} i \cdot 2^i = k \cdot 2^{k+2} +2 $$
for $k \geq 0$

Prove k+1 is true.

$$\sum_{i=1}^{k+2} i \cdot 2^i = (k+1)\cdot 2^{k+3}+2 $$
$\big[RHS\big]$
$k\cdot 2^{k+3}+2^{k+3}+2$
$\big[LHS\big]$
$$\sum_{i=1}^{k+2} {i \cdot 2^{i}} $$
$= \underbrace{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2)\cdot 2^{k+2}$ (Explicit last step)
$= \underbrace{k\cdot 2^{k+2}+2}+(k+2)\cdot 2^{k+2}$ (Inductive Hypothesis Substitution)
$= k\cdot 2^{k+2}+2+k\cdot 2^{k+2}+2^{k+3}$
$= 2k\cdot 2^{k+2} + 2^{k+3} + 2$

My [LHS] has one too many $2k\cdot 2^{k+2}$ or did it just do it completely wrong?
 A: You are done. All you need to do is to regroup the terms properly.
$$2k \cdot 2^{k+2} + 2^{k+3} + 2 = k \cdot 2^{k+3} + 2^{k+3} + 2 = (k+1) \cdot 2^{k+3} + 2$$
which is what you want.
As julien points out, your penultimate step must read $$k \cdot 2^{k+2} + 2 + \color{blue}{k \cdot 2^{k+2}} + 2^{k+3}$$
instead of 
$$k \cdot 2^{k+2} + 2 + \color{red}{k \cdot 2{k+2}} + 2^{k+3}$$
I assume you made a typo while typesetting this line since you missed the ^ symbol i.e. you have 
$$\text{`k \cdot 2{k+2}` instead of `k \cdot 2^{k+2}`}$$
And also as julien points out, your (Explicit last step) should read
$$\color{blue}{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2) \cdot 2^{k+2}$$
and not
$$\color{red}{(k+1) \cdot 2^{k+1}} + (k+2) \cdot 2^{k+2}$$
A: Attention: you replaced $\sum_{i=1}^{k+1}2^i=2+2\cdot 2^2+3\cdot 2^3+\ldots+(k+1)2^{k+1}$ by $(k+1)2^{k+1}$, and this is wrong for $k\geq 2$.
The formula holds for $n=0$ as you observed.
Now assume it holds for some $n\geq 0$, i.e.
$$
\sum_{i=1}^{n+1}i2^i=n2^{n+2}+2.
$$
Then 
$$
\sum_{i=1}^{n+2}i2^i=\sum_{i=1}^{n+1}i2^i+(n+2)2^{n+2}
$$
(so, using the induction hypothesis)
$$
=n2^{n+2}+2+(n+2)2^{n+2}
$$
(so, simplifying)
$$
=2n2^{n+2}+2^{n+3}+2=n2^{n+3}+2^{n+3}+2=(n+1)2^{n+3}+2.
$$
So, by induction, the formula holds for all $n\geq 0$.
