$$ \sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $$
Base Case:
I did $n = 1$, so..
LHS-
$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$
RHS-
$\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} \ = \frac3{8}$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$\sum_{k=1}^{n+1} \frac {k+2}{k(k+1)2^{k+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of $$ \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}} $$ now form here I tried multiplying it out and solving with some algebra but I kept hitting dead ends. If you could explain your steps and the reasoning behind them I would appreciate it.
$$ \begin{split} D &= \frac {n+3}{(n+1)(n+2)2^{n+2}} - \frac1{(n+1)2^{n+1}} \\ &= \frac{1}{(n+1)2^{n+2}} \left[\frac{n+3}{n+2} - 2 \right] \\ &= \frac{1}{(n+1)2^{n+2}} \times \frac{n+3-2n-4}{n+2} \\ &= \frac{1}{(n+1)2^{n+2}} \times \frac{-(n+1)}{n+2} \\ &= \frac{-1}{(n+2)2^{n+2}} \end{split} $$ Can you finish?