`(2)/(1!) + (4)/(3!) + (6)/(5!) + (8)/(7!) + ...oo =`

Options

A. `e^(-1)`
B. e
C. 2e
D. 3e

Answer

To solve the series

\[ S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \ldots \]

we can rewrite the terms in a more manageable form. Notice that the numerators can be expressed as \(2n\) where \(n\) is an integer starting from 1, and the denominators can be expressed as \((2n-1)!\).

Thus, we can rewrite the series as:

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

Next, we can separate the terms in the series:

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

Now, we can express \(n\) in terms of factorials:

\[ n = (2n-1) - (2n-2) \]

This gives us:

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

This can be separated into two sums:

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

The first sum simplifies to:

\[ \sum_{n=1}^{\infty} \frac{1}{(2n-2)!} = \sum_{m=0}^{\infty} \frac{1}{(2m)!} = \cosh(1) \]

The second sum simplifies to:

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

Thus, we can express \(S\) as:

\[ S = 2 \left( \sinh(1) - \cosh(1) \right) \]

Now, we can evaluate \(S\):

\[ S = 2 \left( e - 1 \right) \]

Finally, we find that:

\[ S = 2 \]

Thus, the sum of the series is:

\[ \boxed{2} \]