Let the coefficients of powers of `x` in the `2^(nd), 3^(rd) and 4^(th)` terms in the expansion of `(1 + x)^(n)`, where `n` is a positive integer, be in arithmetic progression. Then the sum of the coefficients of odd powers of `x` in the expansion is

Options

A. 32
B. 64
C. 128
D. None of these

Answer

To solve the problem, we need to find the sum of the coefficients of odd powers of \( x \) in the expansion of \( (1 + x)^n \) given that the coefficients of the 2nd, 3rd, and 4th terms are in arithmetic progression.

### Step-by-Step Solution:

1. **Identify the Terms**:
The expansion of \( (1 + x)^n \) is given by:
\[ T_k = \binom{n}{k} x^k \]
where \( T_k \) is the \( k^{th} \) term. The coefficients of the 2nd, 3rd, and 4th terms are:
- 2nd term: \( \binom{n}{1} \)
- 3rd term: \( \binom{n}{2} \)
- 4th term: \( \binom{n}{3} \)

2. **Set Up the Arithmetic Progression Condition**:
The coefficients are in arithmetic progression, which means:
\[ 2 \cdot \binom{n}{2} = \binom{n}{1} + \binom{n}{3} \]

3. **Substitute the Binomial Coefficients**:
Substitute the binomial coefficients:
\[ 2 \cdot \frac{n!}{2!(n-2)!} = \frac{n!}{1!(n-1)!} + \frac{n!}{3!(n-3)!} \]
Simplifying gives:
\[ \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6} \]

4. **Clear the Denominators**:
Multiply through by 6 to eliminate the fractions:
\[ 3n(n-1) = 6n + n(n-1)(n-2) \]

5. **Rearrange and Simplify**:
Rearranging gives:
\[ 3n^2 - 3n = 6n + n^3 - 3n^2 + 2n \]
Combine like terms:
\[ 0 = n^3 - 6n^2 + 5n \]

6. **Factor the Polynomial**:
Factor out \( n \):
\[ n(n^2 - 6n + 5) = 0 \]
Further factoring gives:
\[ n(n-1)(n-5) = 0 \]
Thus, \( n = 0, 1, 5 \). Since \( n \) is a positive integer, we have \( n = 5 \).

7. **Find the Sum of Coefficients of Odd Powers**:
The sum of the coefficients of odd powers of \( x \) in \( (1 + x)^n \) can be calculated using:
\[ S_{\text{odd}} = \frac{(1 + 1)^n - (1 - 1)^n}{2} \]
Substituting \( n = 5 \):
\[ S_{\text{odd}} = \frac{(2^5) - (0)}{2} = \frac{32}{2} = 16 \]

### Final Answer:
The sum of the coefficients of odd powers of \( x \) in the expansion is \( \boxed{16} \).