`""^1 P_1+2. ""^2 P_2+3.""^3 P_3+....+n. ""^n P_n=`

Options

A. `"^(n+1) P_(n+1)`
B. `"^(n+1) P_(n+1)-1`
C. `"^(n+1) P_(n+1)-2`
D. none of these

Answer

To solve the expression \( S = 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \ldots + n \cdot P_n \), we will follow these steps:

### Step 1: Understand the notation
The notation \( P_n \) refers to permutations of \( n \) objects taken \( r \) at a time, which is given by the formula:
\[ P_n = n! \]
where \( n! \) is the factorial of \( n \).

### Step 2: Rewrite the expression
We can rewrite the expression using the formula for permutations:
\[ S = 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \ldots + n \cdot P_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n! \]

### Step 3: Identify the general term
The general term of the series can be expressed as:
\[ T_n = n \cdot n! \]

### Step 4: Simplify the general term
We can simplify \( T_n \):
\[ T_n = n \cdot n! = (n + 1)! - n! \]
This is derived from the fact that:
\[ (n + 1)! = (n + 1) \cdot n! = n \cdot n! + n! \]

### Step 5: Sum the series
Now we can express the sum \( S \) in terms of the simplified general term:
\[ S = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left( (k + 1)! - k! \right) \]

### Step 6: Recognize the telescoping nature of the series
When we expand this summation, we notice that it is a telescoping series:
\[ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n + 1)! - n!) \]
Most terms will cancel out, leading to:
\[ S = (n + 1)! - 1! \]
Thus, we have:
\[ S = (n + 1)! - 1 \]

### Step 7: Final expression
Therefore, the final expression for the sum is:
\[ S = (n + 1)! - 1 \]