<s>[INST] Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $ 
$$
\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}}
$$

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
$$
\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $\frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$\frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$\frac{2}{(n+1)2^{n+2}} - \frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then
$$\frac{1}2 -\frac{2}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
and
$$\frac{1}2 -\frac{1}{(n+1)2^{n}} - \frac{1}{(n+2)2^{n+2}}$$
$$\frac{1}2 -\frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$\frac{1}2 -\frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$\frac{1}2 -\frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
 [/INST] Your error is just after the sixth step from the bottom:
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
</s>