The sum of the series
`(1)/(19!)+(1)/(3!17!) +(1)/(5!15) + .........` to 10 terms is equal to

Options

A. `(2^(19))/(20!)`
B. `(2^(20))/(20!)`
C. `(2^(10))/(20!)`
D. `(2^(19))/(19!)`

Explanation

To find the sum of the series

\[ S = \frac{1}{19!} + \frac{1}{3! \cdot 17!} + \frac{1}{5! \cdot 15!} + \ldots \]

up to 10 terms, we can follow these steps:

### Step 1: Identify the general term of the series

The series can be expressed in terms of binomial coefficients. The general term can be represented as:

\[ T_n = \frac{1}{(2n-1)! \cdot (19 - 2n)!} \]

for \( n = 1, 2, 3, \ldots, 10 \).

### Step 2: Rewrite the series using binomial coefficients

We can rewrite the series using the property of binomial coefficients:

\[ S = \sum_{n=0}^{9} \frac{1}{(2n)! \cdot (19 - 2n)!} = \frac{1}{20!} \sum_{n=0}^{9} \binom{20}{2n} \]

This is because:

\[ \frac{1}{(2n)! \cdot (19 - 2n)!} = \frac{1}{20!} \cdot \binom{20}{2n} \]

### Step 3: Calculate the sum of binomial coefficients

The sum of the binomial coefficients for even indices can be calculated using the identity:

\[ \sum_{k=0}^{n} \binom{n}{2k} = 2^{n-1} \]

For \( n = 20 \):

\[ \sum_{k=0}^{10} \binom{20}{2k} = 2^{19} \]

### Step 4: Calculate the sum for the first 10 terms

Since we are interested in the first 10 terms (which corresponds to \( n = 0 \) to \( n = 9 \)), we have:

\[ S = \frac{1}{20!} \cdot 2^{19} \]

### Step 5: Final calculation

Now we can compute the final value of \( S \):

\[ S = \frac{2^{19}}{20!} \]

### Conclusion

Thus, the sum of the series up to 10 terms is:

\[ S = \frac{2^{19}}{20!} \]