If `alpha_(1), alpha_(2), …….., alpha_(n)` are the `n,n^(th)` roots of unity, `alpha_(r )=e (i2(r-1)pi)/(n), r=1,2,…n` then `""^(n)C_(1)alpha_(1)+""^(n)C_(2)alpha_(2)+…..+""^(n)C_(n)alpha_(n)` is equal to :

Options

A. `(1+(alpha_(2))/(alpha_(1)))^(n)-1`
B. `(alpha_(1))/(2)[(1+alpha_(1))^(n)-1]`
C. `(alpha_(1)+alpha_(n-1)-1)/(2)`
D. `(alpha_(1)+alpha_(n-1))^(n)-1`

Answer

To solve the problem, we need to evaluate the expression:

\[ \binom{n}{1} \alpha_1 + \binom{n}{2} \alpha_2 + \ldots + \binom{n}{n} \alpha_n \]

where \(\alpha_r = e^{i \frac{2\pi (r-1)}{n}}\) for \(r = 1, 2, \ldots, n\), and \(\alpha_1, \alpha_2, \ldots, \alpha_n\) are the \(n\)-th roots of unity.

### Step 1: Understanding the Roots of Unity
The \(n\)-th roots of unity are given by:
\[ \alpha_r = e^{i \frac{2\pi (r-1)}{n}} \quad \text{for } r = 1, 2, \ldots, n \]
These roots can be represented as:
\[ \alpha_1 = 1, \quad \alpha_2 = e^{i \frac{2\pi}{n}}, \quad \alpha_3 = e^{i \frac{4\pi}{n}}, \ldots, \quad \alpha_n = e^{i \frac{2(n-1)\pi}{n}} \]

### Step 2: Setting Up the Expression
We need to evaluate:
\[ S = \binom{n}{1} \alpha_1 + \binom{n}{2} \alpha_2 + \ldots + \binom{n}{n} \alpha_n \]

### Step 3: Using the Binomial Theorem
We can relate this sum to the binomial expansion. The binomial theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
If we let \(x = \alpha\), we can express our sum \(S\) as:
\[ S = \sum_{r=1}^{n} \binom{n}{r} \alpha_r = \sum_{r=1}^{n} \binom{n}{r} e^{i \frac{2\pi (r-1)}{n}} \]

### Step 4: Adding and Subtracting the Zero Term
We can add and subtract the term for \(r=0\):
\[ S = \left( \sum_{r=0}^{n} \binom{n}{r} \alpha_r \right) - \binom{n}{0} \]
This gives us:
\[ S = (1 + \alpha_1)^n - 1 \]

### Step 5: Evaluating the Final Expression
Now we can evaluate:
\[ S = (1 + 1)^n - 1 = 2^n - 1 \]

### Conclusion
Thus, the value of the expression is:
\[ \boxed{2^n - 1} \]