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 given expression

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

we can follow these steps:

### Step 1: Understand the Binomial Expansion
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: Consider the Expression
We can express the given series in terms of binomial expansions. Notice that the terms with even indices are present in the expansion of \( (1 + x)^{50} \) and \( (1 - x)^{50} \).

### Step 3: Add the Two Expansions
Consider the two expansions:

1. \( (1 + x)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^k \)
2. \( (1 - x)^{50} = \sum_{k=0}^{50} \binom{50}{k} (-x)^k \)

Adding these two expansions cancels out the odd indexed terms:

\[ (1 + x)^{50} + (1 - x)^{50} = 2 \left( \binom{50}{0} + \binom{50}{2} x^2 + \binom{50}{4} x^4 + \ldots \right). \]

### Step 4: Integrate Both Sides
Integrate both sides with respect to \( x \):

\[ \int \left( (1 + x)^{50} + (1 - x)^{50} \right) dx = \int 2 \left( \binom{50}{0} + \binom{50}{2} x^2 + \binom{50}{4} x^4 + \ldots \right) dx. \]

The left-hand side becomes:

\[ \frac{(1 + x)^{51}}{51} - \frac{(1 - x)^{51}}{51} + C. \]

### Step 5: Evaluate at \( x = 1 \)
Now, evaluate both sides at \( x = 1 \):

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

### Step 6: Relate to the Series
The right-hand side evaluates to:

\[ 2 \left( \binom{50}{0} \cdot 1 + \frac{\binom{50}{2}}{3} \cdot 1^3 + \frac{\binom{50}{4}}{5} \cdot 1^5 + \ldots + \frac{\binom{50}{50}}{51} \cdot 1^{51} \right). \]

### Step 7: Solve for the Series
Thus, we have:

\[ \frac{2^{51}}{51} = 2 \left( \frac{\binom{50}{0}}{1} + \frac{\binom{50}{2}}{3} + \frac{\binom{50}{4}}{5} + \ldots + \frac{\binom{50}{50}}{51} \right). \]

Dividing both sides by 2 gives:

\[ \frac{2^{51}}{102} = \frac{\binom{50}{0}}{1} + \frac{\binom{50}{2}}{3} + \frac{\binom{50}{4}}{5} + \ldots + \frac{\binom{50}{50}}{51}. \]

### Final Result
Thus, the value of the expression is:

\[ \frac{2^{51}}{102}. \]