If `1+sum_(r=0)^(18){r(r+2)+1}r!"="k!` then k is ____

Answer

To solve the problem, we need to evaluate the expression given by:

\[ 1 + \sum_{r=0}^{18} \left( r(r+2) + 1 \right) r! \]

We can break this down step by step.

### Step 1: Simplify the summand
First, we simplify the expression inside the summation:

\[ r(r+2) + 1 = r^2 + 2r + 1 = (r + 1)^2 \]

So, we can rewrite the summation as:

\[ \sum_{r=0}^{18} (r + 1)^2 r! \]

### Step 2: Rewrite the summation
Now, we can express \((r + 1)^2\) in terms of factorials:

\[ (r + 1)^2 r! = (r + 1)(r + 1) r! = (r + 1)! (r + 1) \]

Thus, the summation becomes:

\[ \sum_{r=0}^{18} (r + 1)(r + 1)! \]

### Step 3: Change the index of summation
To make it easier, we can change the index of summation by letting \(s = r + 1\). When \(r = 0\), \(s = 1\) and when \(r = 18\), \(s = 19\). Therefore, we rewrite the summation:

\[ \sum_{s=1}^{19} s \cdot s! \]

### Step 4: Factor out the summation
We can factor out \(s \cdot s!\) as follows:

\[ s \cdot s! = (s + 1)! - s! \]

Thus, we can rewrite the summation:

\[ \sum_{s=1}^{19} ((s + 1)! - s!) \]

### Step 5: Evaluate the telescoping series
Now, this is a telescoping series:

\[ = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (20! - 19!) \]

Most terms will cancel out, leaving us with:

\[ 20! - 1! \]

Since \(1! = 1\), we have:

\[ 20! - 1 \]

### Step 6: Add 1 to the result
Now, we go back to our original expression, which was:

\[ 1 + \sum_{r=0}^{18} (r(r+2) + 1) r! = 1 + (20! - 1) = 20! \]

### Step 7: Set the equation equal to \(k!\)
We have:

\[ 20! = k! \]

This implies that \(k = 20\).

### Final Answer
Thus, the value of \(k\) is:

\[ \boxed{20} \]