If ` S = sum_(n=2)^(oo) ""^(n)C_(2) (3^(n-2))/(n!),` then 2S equals

Options

A. `e^(3//2)`
B. `e^(3)`
C. `e^(-3//2)`
D. `e^(-3)`

Answer

To solve the problem, we need to evaluate the sum \( S = \sum_{n=2}^{\infty} \binom{n}{2} \frac{3^{n-2}}{n!} \) and then find \( 2S \).

### Step-by-Step Solution:

1. **Rewrite the Binomial Coefficient**:
The binomial coefficient \( \binom{n}{2} \) can be expressed as:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
Therefore, we can rewrite \( S \):
\[ S = \sum_{n=2}^{\infty} \frac{n(n-1)}{2} \frac{3^{n-2}}{n!} \]

2. **Factor Out the Constant**:
Since \( \frac{1}{2} \) is a constant, we can factor it out of the summation:
\[ S = \frac{1}{2} \sum_{n=2}^{\infty} \frac{n(n-1) 3^{n-2}}{n!} \]

3. **Change the Index of Summation**:
To simplify the summation, we can change the index of summation by letting \( m = n - 2 \). Then \( n = m + 2 \) and when \( n = 2 \), \( m = 0 \):
\[ S = \frac{1}{2} \sum_{m=0}^{\infty} \frac{(m+2)(m+1) 3^{m}}{(m+2)!} \]

4. **Simplify the Factorial**:
We can express \( (m+2)! \) as \( (m+2)(m+1)m! \):
\[ S = \frac{1}{2} \sum_{m=0}^{\infty} \frac{3^{m}}{m!} \]

5. **Recognize the Exponential Series**:
The series \( \sum_{m=0}^{\infty} \frac{3^{m}}{m!} \) is the Taylor series expansion for \( e^x \) evaluated at \( x = 3 \):
\[ \sum_{m=0}^{\infty} \frac{3^{m}}{m!} = e^3 \]

6. **Substitute Back**:
Therefore, we have:
\[ S = \frac{1}{2} e^3 \]

7. **Calculate \( 2S \)**:
Finally, we calculate \( 2S \):
\[ 2S = 2 \cdot \frac{1}{2} e^3 = e^3 \]

### Final Answer:
\[ 2S = e^3 \]