Let's solve the problem step by step using mathematical induction. The task is to prove the following statement for all $n \in \mathbb{N}^+$:

$\sum_{k=1}^n \frac{k+2}{k \cdot (k+1) \cdot 2^k} = 1 - \frac{1}{(n+1) \cdot 2^n}.$

Step 1: Base Case ($n = 1$)

For $n = 1$, the left-hand side of the equation is:

$\sum_{k=1}^1 \frac{k+2}{k \cdot (k+1) \cdot 2^k} = \frac{1+2}{1 \cdot (1+1) \cdot 2^1}.$

Simplify the fraction:

$\frac{1+2}{1 \cdot 2 \cdot 2} = \frac{3}{4}.$

The right-hand side of the equation is:

$1 - \frac{1}{(1+1) \cdot 2^1} = 1 - \frac{1}{2 \cdot 2} = 1 - \frac{1}{4} = \frac{3}{4}.$

Since the left-hand side equals the right-hand side, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for some $n = m$, i.e.,

$\sum_{k=1}^m \frac{k+2}{k \cdot (k+1) \cdot 2^k} = 1 - \frac{1}{(m+1) \cdot 2^m}.$

We need to prove that the statement also holds for $n = m+1$.

Step 3: Inductive Step

For $n = m+1$, the left-hand side becomes:

$\sum_{k=1}^{m+1} \frac{k+2}{k \cdot (k+1) \cdot 2^k}.$

This can be split into two parts:

$\sum_{k=1}^{m+1} \frac{k+2}{k \cdot (k+1) \cdot 2^k} = \sum_{k=1}^m \frac{k+2}{k \cdot (k+1) \cdot 2^k} + \frac{(m+1)+2}{(m+1) \cdot ((m+1)+1) \cdot 2^{m+1}}.$

Using the inductive hypothesis for the first term:

$\sum_{k=1}^m \frac{k+2}{k \cdot (k+1) \cdot 2^k} = 1 - \frac{1}{(m+1) \cdot 2^m}.$

Substitute this back:

$\sum_{k=1}^{m+1} \frac{k+2}{k \cdot (k+1) \cdot 2^k} = 1 - \frac{1}{(m+1) \cdot 2^m} + \frac{(m+1)+2}{(m+1) \cdot ((m+1)+1) \cdot 2^{m+1}}.$

Simplify the second term:

$\frac{(m+1)+2}{(m+1) \cdot ((m+1)+1) \cdot 2^{m+1}} = \frac{m+3}{(m+1) \cdot (m+2) \cdot 2^{m+1}}.$

Now, combine the terms:

$\sum_{k=1}^{m+1} \frac{k+2}{k \cdot (k+1) \cdot 2^k} = 1 + \frac{m+3}{(m+1) \cdot (m+2) \cdot 2^{m+1}} - \frac{1}{(m+1) \cdot 2^m}.$

Rewrite $-\frac{1}{(m+1) \cdot 2^m}$ with a common denominator. Note that:

$\frac{1}{(m+1) \cdot 2^m} = \frac{2}{(m+1) \cdot 2^{m+1}}.$

So the expression becomes:

$1 - \frac{2}{(m+1) \cdot 2^{m+1}} + \frac{m+3}{(m+1) \cdot (m+2) \cdot 2^{m+1}}.$

Combine the fractions under a common denominator, $(m+1) \cdot (m+2) \cdot 2^{m+1}$:

$1 - \frac{2 \cdot (m+2)}{(m+1) \cdot (m+2) \cdot 2^{m+1}} + \frac{m+3}{(m+1) \cdot (m+2) \cdot 2^{m+1}}.$

Simplify the numerator of the combined fraction:

$-(2 \cdot (m+2)) + (m+3) = -2m - 4 + m + 3 = -m - 1.$

So the fraction becomes:

$1 + \frac{-m-1}{(m+1) \cdot (m+2) \cdot 2^{m+1}}.$

Simplify further:

$1 - \frac{1}{(m+2) \cdot 2^{m+1}}.$

This matches the right-hand side for $n = m+1$:

$1 - \frac{1}{(m+2) \cdot 2^{m+1}}.$

Thus, the inductive step is proven.

Conclusion

By mathematical induction, the statement is true for all $n \in \mathbb{N}^+$.

Do you have any questions, or would you like more details about any step? Here are 5 related questions you might consider:

1. What is the importance of the base case in mathematical induction?
2. How do we simplify terms with fractions efficiently during induction proofs?
3. Can this method be used to prove similar summation identities?
4. What are common pitfalls in applying mathematical induction?
5. How can this problem be visualized using partial sums?

Tip: Always double-check your algebraic manipulations during induction to avoid mistakes, especially when dealing with fractions!