Find the value of `1 xx 1!+2 xx 2!+3 xx 3!+........+n xx n!`

Answer

To find the value of the expression \( S = 1 \times 1! + 2 \times 2! + 3 \times 3! + \ldots + n \times n! \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Define the Summation**:
We define the sum \( S \) as follows:
\[ S = \sum_{r=1}^{n} r \times r! \]

2. **Rewrite the General Term**:
We can rewrite the term \( r \times r! \) using the identity \( r = (r + 1) - 1 \):
\[ r \times r! = (r + 1 - 1) \times r! = (r + 1) \times r! - 1 \times r! \]
This simplifies to:
\[ r \times r! = (r + 1)! - r! \]

3. **Substituting Back into the Summation**:
Now, we substitute this back into the summation:
\[ S = \sum_{r=1}^{n} \left( (r + 1)! - r! \right) \]

4. **Expanding the Summation**:
We can expand this summation:
\[ S = \left( 2! - 1! \right) + \left( 3! - 2! \right) + \left( 4! - 3! \right) + \ldots + \left( (n + 1)! - n! \right) \]

5. **Canceling Terms**:
Notice that in this expansion, all the factorial terms cancel out:
\[ S = -1! + (n + 1)! \]
The remaining terms after cancellation are:
\[ S = (n + 1)! - 1 \]

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

### Summary:
The final result for the sum \( 1 \times 1! + 2 \times 2! + 3 \times 3! + \ldots + n \times n! \) is:
\[ S = (n + 1)! - 1 \]