If `|a_(i)| lt 1, lambda_(i) ge 0` for `i=1,2,……n` and `lambda_(1)+lambda_(2)+…….+lambda_(n)=1`, then the value of `|lambda_(1)a_(1)+lambda_(2)a_(2)+…….+lambda_(n)a_(n)|` is

Options

A. equal to 1
B. less than 1
C. greater than 1
D. none of these

Answer

To solve the problem, we need to analyze the given conditions and apply them step by step.

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We have complex numbers \( a_i \) such that \( |a_i| < 1 \) for \( i = 1, 2, \ldots, n \). This means that each \( a_i \) lies within the unit circle in the complex plane. Additionally, we have \( \lambda_i \geq 0 \) and \( \lambda_1 + \lambda_2 + \ldots + \lambda_n = 1 \).

2. **Expressing the Sum**:
We need to evaluate the expression \( | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \). This is a weighted sum of the complex numbers \( a_i \) with weights \( \lambda_i \).

3. **Applying the Triangle Inequality**:
By the triangle inequality, we know that:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \leq | \lambda_1 a_1 | + | \lambda_2 a_2 | + \ldots + | \lambda_n a_n | \]

4. **Calculating Each Term**:
Since \( |a_i| < 1 \), we can write:
\[ | \lambda_i a_i | = \lambda_i |a_i| < \lambda_i \]
Therefore, we have:
\[ | \lambda_1 a_1 | + | \lambda_2 a_2 | + \ldots + | \lambda_n a_n | < \lambda_1 + \lambda_2 + \ldots + \lambda_n = 1 \]

5. **Combining the Results**:
From the triangle inequality and the previous step, we conclude that:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | < 1 \]

6. **Conclusion**:
Since \( | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \) is strictly less than 1, we can state that:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | < 1 \]

### Final Answer:
The value of \( | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \) is less than 1.

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