If `S = sum_(n=2)^(oo) (""^(n)C_(2))/((n+1)!)` then S =

Options

A. `e//2`
B. `e//2 -1`
C. `e//2+1`
D. none

Answer

To solve the problem, we need to evaluate the series:

\[ S = \sum_{n=2}^{\infty} \frac{{n \choose 2}}{{(n+1)!}} \]

### Step 1: Express \( {n \choose 2} \)

The binomial coefficient \( {n \choose 2} \) can be expressed as:

\[ {n \choose 2} = \frac{n(n-1)}{2} \]

### Step 2: Substitute \( {n \choose 2} \) into the series

Substituting this into the series gives:

\[ S = \sum_{n=2}^{\infty} \frac{\frac{n(n-1)}{2}}{(n+1)!} \]

### Step 3: Factor out the constant

We can factor out \( \frac{1}{2} \) from the summation:

\[ S = \frac{1}{2} \sum_{n=2}^{\infty} \frac{n(n-1)}{(n+1)!} \]

### Step 4: Simplify the factorial

Notice that \( (n+1)! = (n+1) \cdot n! \). Thus, we can rewrite the term:

\[ \frac{n(n-1)}{(n+1)!} = \frac{n(n-1)}{(n+1) \cdot n!} \]

### Step 5: Rewrite the series

Now we can rewrite the series as:

\[ S = \frac{1}{2} \sum_{n=2}^{\infty} \frac{n(n-1)}{(n+1) n!} \]

### Step 6: Change the index of summation

To simplify further, we can change the index of summation. Let \( m = n + 1 \), then \( n = m - 1 \), and when \( n = 2 \), \( m = 3 \):

\[ S = \frac{1}{2} \sum_{m=3}^{\infty} \frac{(m-1)(m-2)}{m \cdot (m-1)!} \]

### Step 7: Simplify the expression

This can be simplified as follows:

\[ S = \frac{1}{2} \sum_{m=3}^{\infty} \frac{(m-2)}{(m-1)!} \]

### Step 8: Recognize the series

The series \( \sum_{m=3}^{\infty} \frac{(m-2)}{(m-1)!} \) can be recognized as related to the exponential function \( e^x \):

\[ \sum_{m=0}^{\infty} \frac{m}{m!} = e \]

### Step 9: Evaluate the series

Thus, we can relate our series to \( e \):

\[ S = \frac{1}{2} \cdot e \]

### Final Result

Therefore, the value of \( S \) is:

\[ S = \frac{1}{2} e \]