The value of `((""^(50)C_(0))/(1)+(""^(50)C_(2))/(3)+(""^(50)C_(4))/(5)+….+(""^(50)C_(50))/(51))` is :

Options

A. `2^(50)/51`
B. `(2^(50)-1)/(51)`
C. `(2^(50)-1)/(50)`
D. `(2^(51)-1)/(51)`

Answer

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

\[ \frac{{\binom{50}{0}}}{1} + \frac{{\binom{50}{2}}}{3} + \frac{{\binom{50}{4}}}{5} + \ldots + \frac{{\binom{50}{50}}}{51} \]

### Step 1: Understanding the Binomial Theorem

The Binomial Theorem states that:

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

For \( n = 50 \), we have:

\[ (1 + x)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^k \]

### Step 2: Considering Even and Odd Terms

To separate even and odd terms, we can evaluate \( (1 + x)^{50} \) and \( (1 - x)^{50} \):

\[ (1 - x)^{50} = \sum_{k=0}^{50} \binom{50}{k} (-x)^k = \sum_{k=0}^{50} \binom{50}{k} (-1)^k x^k \]

Adding these two equations:

\[ (1 + x)^{50} + (1 - x)^{50} = 2 \sum_{k \text{ even}} \binom{50}{k} x^k \]

### Step 3: Integrating the Result

Now, we want to find the sum of the even coefficients divided by their respective integers. We can integrate the expression:

\[ \int (1 + x)^{50} \, dx = \frac{(1 + x)^{51}}{51} + C \]

And similarly for \( (1 - x)^{50} \):

\[ \int (1 - x)^{50} \, dx = \frac{(1 - x)^{51}}{51} + C \]

### Step 4: Combining the Integrals

Now we combine the two integrals:

\[ \int (1 + x)^{50} \, dx + \int (1 - x)^{50} \, dx = \frac{(1 + x)^{51}}{51} + \frac{(1 - x)^{51}}{51} \]

### Step 5: Evaluating at \( x = 1 \)

Substituting \( x = 1 \):

\[ \frac{(1 + 1)^{51}}{51} + \frac{(1 - 1)^{51}}{51} = \frac{2^{51}}{51} + 0 = \frac{2^{51}}{51} \]

### Step 6: Final Result

The sum we are looking for is:

\[ \frac{2^{51}}{51} \]

However, since we need to account for the coefficients divided by their respective integers, we can simplify this to:

\[ \frac{2^{50}}{51} \]

Thus, the final answer is:

\[ \frac{2^{50}}{51} \]

### Conclusion

The value of the given expression is:

\[ \frac{2^{50}}{51} \]