<s>[INST] 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?
 [/INST] 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
$$\color{red}{(k+1) \cdot 2^{k+1}} + (k+2) \cdot 2^{k+2}$$
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