If `nge3and1,alpha_(1),alpha_(2),.......,alpha_(n-1)` are nth roots of unity then the sum `sum_(1leiltjlen-1)alpha_(i)alpha(j)=`

Answer

To solve the problem, we need to find the sum \( \sum_{1 \leq i \leq j \leq n-1} \alpha_i \alpha_j \), where \( \alpha_1, \alpha_2, \ldots, \alpha_{n-1} \) are the \( n \)-th roots of unity.

### Step-by-Step Solution:

1. **Understanding the \( n \)-th Roots of Unity**:
The \( n \)-th roots of unity are given by the formula:
\[ \alpha_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
However, since we are summing from \( 1 \) to \( n-1 \), we will consider \( \alpha_1, \alpha_2, \ldots, \alpha_{n-1} \).

2. **Sum of Roots**:
The sum of all \( n \)-th roots of unity is:
\[ \sum_{k=0}^{n-1} \alpha_k = 0 \]
This implies:
\[ 1 + \sum_{k=1}^{n-1} \alpha_k = 0 \Rightarrow \sum_{k=1}^{n-1} \alpha_k = -1 \]

3. **Calculating the Required Sum**:
We need to calculate:
\[ \sum_{1 \leq i \leq j \leq n-1} \alpha_i \alpha_j \]
This can be rewritten using the identity:
\[ \sum_{i=1}^{n-1} \alpha_i^2 + 2 \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j \]
Let \( S = \sum_{i=1}^{n-1} \alpha_i \). Then:
\[ S^2 = \left(\sum_{i=1}^{n-1} \alpha_i\right)^2 = \sum_{i=1}^{n-1} \alpha_i^2 + 2 \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j \]
Rearranging gives:
\[ 2 \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = S^2 - \sum_{i=1}^{n-1} \alpha_i^2 \]

4. **Finding \( S^2 \)**:
Since \( S = -1 \):
\[ S^2 = (-1)^2 = 1 \]

5. **Finding \( \sum_{i=1}^{n-1} \alpha_i^2 \)**:
The sum of the squares of the \( n \)-th roots of unity can be calculated as follows:
\[ \sum_{i=1}^{n-1} \alpha_i^2 = \sum_{k=0}^{n-1} \alpha_k^2 - 1 = \sum_{k=0}^{n-1} e^{4\pi i k/n} - 1 \]
The sum \( \sum_{k=0}^{n-1} e^{4\pi i k/n} \) is also zero since it represents the \( n \)-th roots of unity for \( n \) roots. Thus:
\[ \sum_{i=1}^{n-1} \alpha_i^2 = 0 - 1 = -1 \]

6. **Final Calculation**:
Substituting back into our equation:
\[ 2 \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 1 - (-1) = 2 \]
Therefore:
\[ \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 1 \]

7. **Conclusion**:
The required sum is:
\[ \sum_{1 \leq i \leq j \leq n-1} \alpha_i \alpha_j = \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j + \sum_{i=1}^{n-1} \alpha_i^2 = 1 + (-1) = 0 \]

### Final Answer:
\[ \sum_{1 \leq i \leq j \leq n-1} \alpha_i \alpha_j = 0 \]