Find the following sum: `1/(n !)+1/(2!(n-2)!)+1/(4!(n-4)!)+. . . .`

Explanation

To find the sum

\[ S = \frac{1}{n!} + \frac{1}{2!(n-2)!} + \frac{1}{4!(n-4)!} + \ldots \]

we can follow these steps:

### Step 1: Rewrite the terms
We can express the terms in a more manageable form. Notice that each term can be rewritten as:

\[ \frac{1}{k!(n-k)!} \]

for \( k = 0, 2, 4, \ldots \). Thus, we can rewrite our sum as:

\[ S = \sum_{k=0, k \text{ even}}^{n} \frac{1}{k!(n-k)!} \]

### Step 2: Multiply and divide by \( n! \)
To facilitate the computation, we multiply and divide the entire sum by \( n! \):

\[ S = \frac{1}{n!} \left( n! + \frac{n!}{2!(n-2)!} + \frac{n!}{4!(n-4)!} + \ldots \right) \]

### Step 3: Recognize the binomial coefficients
The terms inside the parentheses can be recognized as binomial coefficients:

\[ S = \frac{1}{n!} \left( \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \ldots \right) \]

### Step 4: Use the binomial theorem
The sum of the coefficients in the binomial expansion of \( (1 + 1)^n \) gives us \( 2^n \). However, since we are only summing the coefficients for even indices, we can use the identity:

\[ \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \ldots = \frac{1}{2} \cdot 2^n = 2^{n-1} \]

### Step 5: Substitute back into the equation
Now substituting this back into our expression for \( S \):

\[ S = \frac{1}{n!} \cdot 2^{n-1} \]

### Final Result
Thus, the required sum is:

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