If `C_0, C_1,C_2 ..., C_n`, denote the binomial coefficients in the expansion of `(1 + x)^n`, then `C_1/2+C_3/4+C_5/6+......` is equal to

Options

A. `(2^(n) -1)/(n+1)`
B. `(2^(n))/(n+2)`
C. `(2^(n-1))/(n)`
D. `(2^(n-1))/(n+1)`

Explanation

To solve the problem of finding the sum \( S = \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots \), where \( C_k \) are the binomial coefficients from the expansion of \( (1 + x)^n \), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \( (1 + x)^n \) can be expressed as:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \).

### Step 2: Consider the Expansion of \( (1 - x)^n \)
Similarly, the expansion of \( (1 - x)^n \) is:
\[ (1 - x)^n = C_0 - C_1 x + C_2 x^2 - C_3 x^3 + \ldots + (-1)^n C_n x^n \]

### Step 3: Subtract the Two Expansions
Now, if we subtract the second expansion from the first:
\[ (1 + x)^n - (1 - x)^n = (C_1 + C_3 x^3 + C_5 x^5 + \ldots) - (C_1 - C_3 x^3 + C_5 x^5 - \ldots) \]
This simplifies to:
\[ 2(C_1 x + C_3 x^3 + C_5 x^5 + \ldots) \]

### Step 4: Simplify the Expression
The left-hand side becomes:
\[ (1 + x)^n - (1 - x)^n = 2 \sum_{k \text{ odd}} C_k x^k \]
Thus, we can express the sum of odd-indexed coefficients as:
\[ \sum_{k \text{ odd}} C_k x^k = \frac{(1 + x)^n - (1 - x)^n}{2} \]

### Step 5: Integrate to Find the Required Sum
To find the desired sum \( S \), we can integrate the expression:
\[ S = \int_0^1 \left( \sum_{k \text{ odd}} C_k x^k \right) dx \]
This gives:
\[ S = \int_0^1 \frac{(1 + x)^n - (1 - x)^n}{2} dx \]

### Step 6: Evaluate the Integral
Calculating the integral:
\[ S = \frac{1}{2} \left[ \frac{(1 + x)^{n+1}}{n+1} - \frac{(1 - x)^{n+1}}{n+1} \right]_0^1 \]
Evaluating at the limits:
- At \( x = 1 \): \( (1 + 1)^{n+1} = 2^{n+1} \) and \( (1 - 1)^{n+1} = 0 \)
- At \( x = 0 \): both terms equal to \( 1 \)

Thus:
\[ S = \frac{1}{2} \left( \frac{2^{n+1}}{n+1} - 0 \right) = \frac{2^{n+1}}{2(n+1)} = \frac{2^n}{n+1} \]

### Final Answer
Therefore, the final result is:
\[ S = \frac{2^n}{n + 1} \]