If `a_n=n/((n+1)!)` then find `sum_(n=1)^50 a_n`

Options

A. `(50!-1)/(50!)`
B. `(51!-1)/(51!)`
C. `(1)/(2(n-1)!)`
D. none of these

Explanation

To solve the problem, we need to find the sum \( S = \sum_{n=1}^{50} a_n \) where \( a_n = \frac{n}{(n+1)!} \).

### Step-by-Step Solution:

1. **Rewrite \( a_n \)**:
We start with the expression for \( a_n \):
\[ a_n = \frac{n}{(n+1)!} \]
We can rewrite \( n \) as \( (n+1) - 1 \):
\[ a_n = \frac{(n+1) - 1}{(n+1)!} = \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!} \]
This simplifies to:
\[ a_n = \frac{1}{n!} - \frac{1}{(n+1)!} \]

2. **Set up the sum**:
Now we can express the sum \( S \):
\[ S = \sum_{n=1}^{50} a_n = \sum_{n=1}^{50} \left( \frac{1}{n!} - \frac{1}{(n+1)!} \right) \]

3. **Recognize the telescoping series**:
The series is telescoping. When we expand it, we get:
\[ S = \left( \frac{1}{1!} - \frac{1}{2!} \right) + \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) + \ldots + \left( \frac{1}{50!} - \frac{1}{51!} \right) \]
Most terms cancel out, leaving us with:
\[ S = \frac{1}{1!} - \frac{1}{51!} \]

4. **Calculate the final result**:
We know that \( \frac{1}{1!} = 1 \), so we have:
\[ S = 1 - \frac{1}{51!} \]

### Final Answer:
Thus, the sum \( \sum_{n=1}^{50} a_n \) is:
\[ S = 1 - \frac{1}{51!} \]