In the expansion of `( 1+ x)^(50)`, the sum of the coefficient of odd powers of x is `:`

Options

A. `2^(26)`
B. `2^(49)`
C. `2^(50)`
D. `2^(51)`

Answer

To find the sum of the coefficients of the odd powers of \( x \) in the expansion of \( (1 + x)^{50} \), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \( (1 + x)^{n} \) is given by:
\[ (1 + x)^{n} = \sum_{k=0}^{n} C(n, k) x^{k} \]
where \( C(n, k) \) is the binomial coefficient, which represents the coefficient of \( x^k \).

### Step 2: Identify the Total Sum of Coefficients
The sum of all coefficients in the expansion of \( (1 + x)^{50} \) can be found by substituting \( x = 1 \):
\[ (1 + 1)^{50} = 2^{50} \]
This means the total sum of all coefficients is \( 2^{50} \).

### Step 3: Separate Even and Odd Coefficients
To find the sum of the coefficients of odd powers of \( x \), we can use the property that:
\[ \text{Sum of coefficients of odd powers} = \frac{1}{2} \left( \text{Total sum of coefficients} + \text{Sum of coefficients of even powers} \right) \]
However, a simpler approach is to use the fact that:
\[ \text{Sum of coefficients of odd powers} = \frac{(1 + 1)^{50} - (1 - 1)^{50}}{2} \]
Here, \( (1 - 1)^{50} = 0 \).

### Step 4: Calculate the Sum of Odd Coefficients
Using the above formula:
\[ \text{Sum of coefficients of odd powers} = \frac{2^{50} - 0}{2} = \frac{2^{50}}{2} = 2^{49} \]

### Conclusion
Thus, the sum of the coefficients of the odd powers of \( x \) in the expansion of \( (1 + x)^{50} \) is:
\[ \boxed{2^{49}} \]