If`S=Sigma_(n=1)^(oo) (""^(n)C_(0)+""^(n)C_(1)+""^(n)c_(2)+..+""^(n)C_(n))/(""^(n)P_(n))` then S equals

Options

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

Answer

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

\[ S = \sum_{n=1}^{\infty} \frac{\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}}{P(n, n)} \]

### Step 1: Understanding the components of the series

The numerator of the series consists of the sum of binomial coefficients, which can be simplified. The sum of the binomial coefficients for a given \( n \) is:

\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]

The denominator, \( P(n, n) \), represents the number of permutations of \( n \) items taken \( n \) at a time, which is equal to \( n! \).

### Step 2: Rewrite the series

Now we can rewrite the series \( S \) as:

\[ S = \sum_{n=1}^{\infty} \frac{2^n}{n!} \]

### Step 3: Recognizing the series as an exponential function

The series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is known to converge to \( e^x \). Therefore, we can express our series as:

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

Here, we subtract 1 because the series starts from \( n=1 \) instead of \( n=0 \).

### Step 4: Conclusion

Thus, the final result for \( S \) is:

\[ S = e^2 - 1 \]

### Final Answer

The value of \( S \) is \( e^2 - 1 \).

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