The sum of the series
`""^(3)C_(0)- ""^(4)C_(1) . (1)/(2) + ""^(5)C_(2)((1)/(2))^(2) - ""^(6)C_(3)((1)/(2))^(3) + ...` to `infty`, is

Options

A. 16
B. 8
C. `(16)/(81)`
D. none of these

Explanation

To find the sum of the series

\[ S = \sum_{n=0}^{\infty} (-1)^n \binom{n+3}{3} \left(\frac{1}{2}\right)^n \]

we will follow a systematic approach.

### Step 1: Identify the series
The series can be rewritten as:

\[ S = \binom{3}{0} - \binom{4}{1} \left(\frac{1}{2}\right) + \binom{5}{2} \left(\frac{1}{2}\right)^2 - \binom{6}{3} \left(\frac{1}{2}\right)^3 + \ldots \]

### Step 2: Use the Binomial Theorem
We can relate this series to the binomial theorem. The binomial theorem states that:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

In our case, we need to consider the negative sign and the terms of the series.

### Step 3: Transform the series
We can express the series in terms of a generating function. The series can be represented as:

\[ S = \sum_{n=0}^{\infty} \binom{n+3}{3} (-1)^n \left(\frac{1}{2}\right)^n \]

### Step 4: Recognize the generating function
The generating function for the series \(\sum_{n=0}^{\infty} \binom{n+k}{k} x^n\) is given by:

\[ \frac{1}{(1-x)^{k+1}} \]

For our series, \(k = 3\) and \(x = -\frac{1}{2}\):

\[ \sum_{n=0}^{\infty} \binom{n+3}{3} \left(-\frac{1}{2}\right)^n = \frac{1}{\left(1 + \frac{1}{2}\right)^{4}} = \frac{1}{\left(\frac{3}{2}\right)^{4}} = \frac{16}{81} \]

### Step 5: Final result
Thus, the sum of the series is:

\[ S = \frac{16}{81} \]

### Conclusion
The sum of the series is:

\[ \boxed{\frac{16}{81}} \]