Let `t_n =n.(n!)` Then `sum_(n=1)^(15) t_n` is equal to

Options

A. 15!-1
B. 15!+1
C. 16!-1
D. 16!+1

Answer

To solve the problem, we need to find the sum \( S = \sum_{n=1}^{15} t_n \) where \( t_n = n \cdot n! \).

### Step-by-Step Solution:

1. **Define the term**:
\[ t_n = n \cdot n! \]

2. **Write out the sum**:
\[ S = \sum_{n=1}^{15} t_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + 15 \cdot 15! \]

3. **Express \( n \cdot n! \)**:
Notice that \( n \cdot n! = (n+1)! - n! \). This can be derived from:
\[ n \cdot n! = n! + n! = (n+1)! - n! \]

4. **Rewrite the sum using the new expression**:
\[ S = \sum_{n=1}^{15} ((n+1)! - n!) \]

5. **Expand the sum**:
\[ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (16! - 15!) \]

6. **Observe the cancellation**:
In the expanded sum, all terms cancel except for the last term of the last factorial and the first term of the first factorial:
\[ S = 16! - 1! \]

7. **Simplify**:
Since \( 1! = 1 \), we have:
\[ S = 16! - 1 \]

### Final Answer:
Thus, the value of \( \sum_{n=1}^{15} t_n \) is:
\[ \boxed{16! - 1} \]