# 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 $$2^{n}$$

$$\frac{1}2 -\frac{(3n-7)}{(n+1)(n+2)2^{n+2}}$$

$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$
Using a telescoping sum, we get \begin{align} \sum_{k=1}^n\frac{k+2}{k(k+1)2^{k+1}} &=\sum_{k=1}^n\left(\frac1{k2^k}-\frac1{(k+1)2^{k+1}}\right)\\ &=\sum_{k=1}^n\frac1{k2^k}-\sum_{k=2}^{n+1}\frac1{k2^k}\\ &=\frac12-\frac1{(n+1)2^{n+1}} \end{align}