If `alpha !=1` is an `n^(th)` root of unity and `n in N` such thatfirst three terms in the expansion of `(alpha + x)^n` are `1, alpha and (n - 1)/(2n) bar a^2`, then the value of x, is

Options

A. `(1)//(n)`
B. `(2)//(n)`
C. `1//2`
D. `(1)//(4)`

Explanation

To solve the problem step by step, we will analyze the given information and use the Binomial Theorem to find the value of \( x \).

### Step 1: Understanding the Problem
We know that \( \alpha \) is an \( n^{th} \) root of unity, which means \( \alpha^n = 1 \). We are given that the first three terms in the expansion of \( (\alpha + x)^n \) are \( 1, \alpha, \) and \( \frac{n-1}{2n} \alpha^2 \).

### Step 2: Applying the Binomial Theorem
According to the Binomial Theorem, the expansion of \( (\alpha + x)^n \) is:
\[ (\alpha + x)^n = \sum_{k=0}^{n} \binom{n}{k} \alpha^{n-k} x^k \]
The first three terms of this expansion are:
1. \( \binom{n}{0} \alpha^n = 1 \)
2. \( \binom{n}{1} \alpha^{n-1} x = n \alpha^{n-1} x \)
3. \( \binom{n}{2} \alpha^{n-2} x^2 = \frac{n(n-1)}{2} \alpha^{n-2} x^2 \)

### Step 3: Setting Up the Equations
From the problem, we have:
1. The first term \( 1 \) is already satisfied because \( \alpha^n = 1 \).
2. The second term gives us the equation:
\[ n \alpha^{n-1} x = \alpha \]
Rearranging gives:
\[ x = \frac{\alpha}{n \alpha^{n-1}} = \frac{1}{n} \quad \text{(since } \alpha^n = 1\text{)} \]

### Step 4: Using the Third Term
The third term gives us the equation:
\[ \frac{n(n-1)}{2} \alpha^{n-2} x^2 = \frac{n-1}{2n} \alpha^2 \]
Substituting \( x = \frac{1}{n} \):
\[ \frac{n(n-1)}{2} \alpha^{n-2} \left(\frac{1}{n}\right)^2 = \frac{n-1}{2n} \alpha^2 \]
This simplifies to:
\[ \frac{(n-1)}{2n} \alpha^{n-2} = \frac{n-1}{2n} \alpha^2 \]

### Step 5: Equating the Terms
Since \( \frac{n-1}{2n} \) is common on both sides, we can cancel it out (assuming \( n \neq 1 \)):
\[ \alpha^{n-2} = \alpha^2 \]
This leads to:
\[ \alpha^{n-4} = 1 \]
Since \( \alpha \) is an \( n^{th} \) root of unity, this is satisfied.

### Final Result
Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{n}} \]