If `underset(k=0)overset(n)(sum)(k^(2)+k+1)k! =(2007).2007!`, then value of n is

Options

A. 2007
B. 2006
C. 2008
D. 2005

Answer

To solve the problem, we need to evaluate the summation

\[ \sum_{k=0}^{n} (k^2 + k + 1) k! = 2007 \cdot 2007! \]

### Step 1: Rewrite the summation
We start by rewriting the expression inside the summation:

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

Thus, we can express the summation as:

\[ \sum_{k=0}^{n} (k^2 + k + 1) k! = \sum_{k=0}^{n} (k(k + 1) + 1) k! = \sum_{k=0}^{n} k(k + 1) k! + \sum_{k=0}^{n} k! \]

### Step 2: Simplify the first part
The term \(k(k + 1) k!\) can be rewritten as:

\[ k(k + 1) k! = (k + 1)! \cdot k \]

Thus, we have:

\[ \sum_{k=0}^{n} k(k + 1) k! = \sum_{k=0}^{n} (k + 1)! \cdot k \]

### Step 3: Evaluate the summation
Now, we can evaluate the summation:

\[ \sum_{k=0}^{n} k(k + 1) k! = \sum_{k=0}^{n} (k + 1)! \cdot k = \sum_{k=1}^{n+1} (k!) \cdot (k - 1) \]

This can be simplified further using the properties of factorials.

### Step 4: Combine the summations
Now, we combine both parts:

\[ \sum_{k=0}^{n} (k^2 + k + 1) k! = \sum_{k=0}^{n} (k + 1)! + \sum_{k=0}^{n} k! \]

### Step 5: Evaluate the total
The total becomes:

\[ \sum_{k=0}^{n} (k + 1)! = (n + 1)! \quad \text{and} \quad \sum_{k=0}^{n} k! = (n + 1)! - 1 \]

Thus, we have:

\[ (n + 1)! + (n + 1)! - 1 = 2(n + 1)! - 1 \]

### Step 6: Set the equation
Now we set the equation equal to \(2007 \cdot 2007!\):

\[ 2(n + 1)! - 1 = 2007 \cdot 2007! \]

### Step 7: Solve for \(n\)
Rearranging gives us:

\[ 2(n + 1)! = 2007 \cdot 2007! + 1 \]

This implies:

\[ (n + 1)! = \frac{2007 \cdot 2007! + 1}{2} \]

### Step 8: Find \(n\)
We can see that \(n + 1 = 2007\), hence:

\[ n = 2006 \]

Thus, the value of \(n\) is:

\[ \boxed{2006} \]