In the expansion of `(1+x)^(50),` find the sum of coefficients of odd powers of `x`.

Explanation

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:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For our case, \( n = 50 \), so:
\[ (1+x)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^k \]

### Step 2: Identify Coefficients of Odd Powers
We need to find the sum of the coefficients of odd powers of \( x \). The odd powers in this expansion are \( x^1, x^3, x^5, \ldots, x^{49} \). The coefficients corresponding to these powers are \( \binom{50}{1}, \binom{50}{3}, \binom{50}{5}, \ldots, \binom{50}{49} \).

### Step 3: Use the Binomial Theorem with \( x = 1 \) and \( x = -1 \)
To find the sum of the coefficients of odd powers, we can evaluate the polynomial at \( x = 1 \) and \( x = -1 \).

1. **Evaluate at \( x = 1 \)**:
\[ (1 + 1)^{50} = 2^{50} \]
This gives us the sum of all coefficients:
\[ \sum_{k=0}^{50} \binom{50}{k} = 2^{50} \]

2. **Evaluate at \( x = -1 \)**:
\[ (1 - 1)^{50} = 0 \]
This gives us the alternating sum of the coefficients:
\[ \sum_{k=0}^{50} (-1)^k \binom{50}{k} = 0 \]

### Step 4: Set Up the Equations
From the evaluations:
- Let \( S \) be the sum of the coefficients of odd powers:
\[ S = \binom{50}{1} + \binom{50}{3} + \binom{50}{5} + \ldots + \binom{50}{49} \]
- Let \( T \) be the sum of the coefficients of even powers:
\[ T = \binom{50}{0} + \binom{50}{2} + \binom{50}{4} + \ldots + \binom{50}{50} \]

From our previous evaluations, we have:
1. \( S + T = 2^{50} \)
2. \( T - S = 0 \)

### Step 5: Solve the Equations
From \( T - S = 0 \), we can deduce that:
\[ T = S \]
Substituting \( T \) in the first equation gives:
\[ S + S = 2^{50} \implies 2S = 2^{50} \implies S = 2^{49} \]

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