If `P_r` stands for `"^rP_r`, then the value of
`1+1.P_1 +2.P_2+3.P_3+....+n.P_n` is :

Options

A. `(n-1)!/2`
B. `(n+1)!/2`
C. `2(n-1)!`
D. `(n+1)!`

Answer

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

\[ S = 1 + 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \ldots + n \cdot P_n \]

where \( P_r \) is defined as \( r! \) (the factorial of \( r \)).

### Step-by-Step Solution:

1. **Understanding \( P_r \)**:
\[ P_r = r! \]
Therefore, we can rewrite the expression \( S \) as:
\[ S = 1 + 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n! \]

2. **Expanding the Terms**:
The expression can be expanded as follows:
\[ S = 1 + 1 + 2 \cdot 2 + 3 \cdot 6 + 4 \cdot 24 + \ldots + n \cdot n! \]

3. **Rewriting Each Term**:
Notice that \( k \cdot k! = (k + 1)! - k! \). This means:
\[ k \cdot k! = (k + 1)! - k! \]
Thus, we can rewrite the sum:
\[ S = 1 + \sum_{k=1}^{n} ((k + 1)! - k!) \]

4. **Simplifying the Sum**:
The sum can be simplified:
\[ S = 1 + \left( (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n + 1)! - n!) \right) \]
Notice that this is a telescoping series, where most terms cancel out:
\[ S = 1 + (n + 1)! - 1! \]
Since \( 1! = 1 \), we have:
\[ S = (n + 1)! + 1 - 1 = (n + 1)! \]

5. **Final Result**:
Therefore, the final value of the expression is:
\[ S = (n + 1)! \]

### Conclusion:
The value of \( 1 + 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \ldots + n \cdot P_n \) is \( (n + 1)! \).