1.1!+2.2!+3.3!+.......n.n!` is equal to

Options

A. `(n+1)!`
B. `(n+1)!+1`
C. `(n+1)!-1`
D. none of these

Explanation

To solve the problem \(1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n!\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ S = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n! \]
We can express this as a summation:
\[ S = \sum_{r=1}^{n} r \cdot r! \]

**Hint:** Recognize that \(r \cdot r!\) can be manipulated to reveal a pattern.

### Step 2: Use the identity
We know that:
\[ r \cdot r! = (r + 1)! - r! \]
This means we can rewrite our summation:
\[ S = \sum_{r=1}^{n} ((r + 1)! - r!) \]

**Hint:** Look for a telescoping series in the summation.

### Step 3: Expand the summation
Now, let's expand the summation:
\[ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n + 1)! - n!) \]
Notice that this is a telescoping series where most terms cancel out.

**Hint:** Identify the first and last terms that do not cancel.

### Step 4: Simplify the expression
After cancellation, we are left with:
\[ S = (n + 1)! - 1! \]
Since \(1! = 1\), we can simplify this to:
\[ S = (n + 1)! - 1 \]

**Hint:** Ensure you substitute \(1!\) correctly to finalize your answer.

### Final Answer
Thus, the final result for the expression \(1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n!\) is:
\[ S = (n + 1)! - 1 \]

### Summary of Steps
1. Rewrite the expression as a summation.
2. Use the identity \(r \cdot r! = (r + 1)! - r!\).
3. Expand the summation to see the telescoping nature.
4. Simplify the remaining terms to get the final answer.