` sum_(n=1)^(oo) (""^(n)C_(0) + ""^(n)C_(1) + .......""^(n)C_(n))/(n!)` is equal to

Options

A. `e^(2)`
B. `e^(2) + 1`
C. `e^(2) - 1`
D. `e^(-2)`

Answer

To solve the problem, we need to evaluate the infinite series given by:

\[ \sum_{n=1}^{\infty} \frac{\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n}}{n!} \]

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The sum of the binomial coefficients for a given \( n \) is given by:
\[ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n \]
This is because the sum of the coefficients in the binomial expansion of \( (1 + 1)^n \) equals \( 2^n \).

2. **Rewriting the Series**:
We can rewrite the series as:
\[ \sum_{n=1}^{\infty} \frac{2^n}{n!} \]

3. **Recognizing the Exponential Series**:
The series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is the Taylor series expansion for \( e^x \). Therefore, we can express our series as:
\[ \sum_{n=0}^{\infty} \frac{2^n}{n!} - \frac{2^0}{0!} \]
This accounts for the \( n=0 \) term, which we do not want in our original series.

4. **Calculating the Exponential Series**:
Thus, we have:
\[ e^2 - 1 \]

5. **Final Result**:
Therefore, the value of the original series is:
\[ e^2 - 1 \]

### Conclusion:
The final answer for the given series is:
\[ e^2 - 1 \]