Let the coefficients of powers of 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. The sum of the coefficients of odd powers of x in the expansion is

Options

A. 32
B. 64
C. 128
D. 256

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 second, third, and fourth terms are in arithmetic progression. Let's break this down step by step.

### Step 1: Identify the coefficients of the terms
The coefficients of the terms in the expansion of \( (1+x)^n \) can be expressed using binomial coefficients:
- The coefficient of the second term is \( \binom{n}{1} \)
- The coefficient of the third term is \( \binom{n}{2} \)
- The coefficient of the fourth term is \( \binom{n}{3} \)

### Step 2: Set up the arithmetic progression condition
For these coefficients to be in arithmetic progression, the following condition must hold:
\[ 2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3} \]

### Step 3: Substitute the binomial coefficients
Substituting the binomial coefficients into the equation, we have:
\[ 2 \cdot \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6} \]
This simplifies to:
\[ n(n-1) = n + \frac{n(n-1)(n-2)}{6} \]

### Step 4: Multiply through by 6 to eliminate the fraction
Multiplying the entire equation by 6 gives:
\[ 6n(n-1) = 6n + n(n-1)(n-2) \]

### Step 5: Expand and rearrange the equation
Expanding both sides:
\[ 6n^2 - 6n = 6n + n^3 - 3n^2 + 2n \]
Rearranging gives:
\[ n^3 - 9n^2 + 12n = 0 \]

### Step 6: Factor the polynomial
Factoring out \( n \):
\[ n(n^2 - 9n + 12) = 0 \]
Now we solve the quadratic:
\[ n^2 - 9n + 12 = 0 \]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 48}}{2} = \frac{9 \pm \sqrt{33}}{2} \]

### Step 7: Determine valid integer solutions
Since \( n \) must be a positive integer, we check for integer solutions. The only integer solution from our factored form is \( n = 7 \) (as \( n = 2 \) is not valid for having four terms).

### Step 8: Calculate the sum of coefficients of odd powers of \( x \)
The sum of the coefficients of odd powers of \( x \) in the expansion of \( (1+x)^n \) is given by:
\[ \frac{(1+1)^n - (1-1)^n}{2} = \frac{2^n - 0}{2} = 2^{n-1} \]
Substituting \( n = 7 \):
\[ \text{Sum} = 2^{7-1} = 2^6 = 64 \]

### Final Answer
The sum of the coefficients of odd powers of \( x \) in the expansion of \( (1+x)^7 \) is \( \boxed{64} \).