The given expression is:

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

We will simplify each term and try to find a pattern that can lead to a more efficient calculation.

Step 1: Simplifying the denominator

The denominator of each term is a sum of factorials:

$k! + (k+1)! + (k+2)!.$

We can factor out $k!$ from each term:

$k! + (k+1)! + (k+2)! = k! + (k+1)k! + (k+2)(k+1)k!.$

This simplifies to:

$k!(1 + (k+1) + (k+2)(k+1)).$

Expanding the terms in parentheses:

$1 + (k+1) + (k+2)(k+1) = 1 + k + 1 + k^2 + 3k + 2 = k^2 + 4k + 4.$

Thus, the denominator becomes:

$k!(k^2 + 4k + 4).$

Step 2: Simplifying each term in the series

Now, we look at each term in the sum:

$\frac{k+2}{k!(k^2 + 4k + 4)}.$

Notice that:

$k^2 + 4k + 4 = (k+2)^2,$

so the denominator becomes:

$k!(k+2)^2.$

Thus, each term of the sum can be rewritten as:

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

Step 3: Final expression for the sum

So, the original sum becomes:

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

This is the simplified expression for the given series.

Step 4: Generalizing the result

While we have simplified the series, evaluating it further requires summing the individual terms for specific values of $n$. However, if needed, each term can be computed by plugging in the desired $k$ and summing over the range from 1 to $n$.

Let me know if you need further details or if you'd like to explore other aspects of the problem.

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Related Questions:

1. How can I compute the sum for specific values of $n$?
2. What happens if we extend the series to infinity?
3. Can we generalize this type of factorial sum for other similar expressions?
4. What is the convergence behavior of factorial-based sums like this?
5. How does the choice of $k$ in the denominator affect the series?

Tip: Factoring out common terms in factorial-based sums often simplifies the expressions and reveals patterns that are easier to work with.