The value of `""^(2)P_(1)+""^(3)P_(1)+……+ ""^(n)P_(1)` is equal to :

Options

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

Answer

To solve the problem \( 2P_1 + 3P_1 + \ldots + nP_1 \), we will follow these steps:

### Step 1: Understand the notation
The notation \( nP_r \) (or \( nPr \)) represents the number of permutations of \( n \) items taken \( r \) at a time. The formula for \( nP_r \) is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]

### Step 2: Apply the formula for \( r = 1 \)
For \( r = 1 \), the formula simplifies to:
\[ nP_1 = \frac{n!}{(n-1)!} = n \]
Thus, we can rewrite the expression:
\[ 2P_1 + 3P_1 + \ldots + nP_1 = 2 + 3 + 4 + \ldots + n \]

### Step 3: Rewrite the sum
The expression \( 2 + 3 + 4 + \ldots + n \) can be rewritten as:
\[ (1 + 2 + 3 + \ldots + n) - 1 \]
This is because we are starting our sum from 2 instead of 1.

### Step 4: Use the formula for the sum of the first \( n \) natural numbers
The formula for the sum of the first \( n \) natural numbers is:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, we can substitute this into our expression:
\[ 2 + 3 + 4 + \ldots + n = \frac{n(n + 1)}{2} - 1 \]

### Step 5: Final expression
Putting it all together, we find:
\[ 2P_1 + 3P_1 + \ldots + nP_1 = \frac{n(n + 1)}{2} - 1 \]

### Conclusion
The value of \( 2P_1 + 3P_1 + \ldots + nP_1 \) is:
\[ \frac{n(n + 1)}{2} - 1 \]

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