If `|a_(i)|lt1lamda_(i)ge0` for `i=1,2,3,.......nandlamda_(1)+lamda_(2)+.......+lamda_(n)=1` then the value of `|lamda_(1)a_(1)+lamda_(2)a_(2)+.......+lamda_(n)a_(n)|` is :

Options

A. Equal to 1
B. Less then 1
C. Greater then 1
D. None of these

Answer

To solve the problem, we need to find the value of \( | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \) under the given conditions. Let's break it down step by step.

### Step 1: Understand the given conditions
We know that:
1. \( |a_i| < 1 \) for \( i = 1, 2, \ldots, n \)
2. \( \lambda_i \geq 0 \) for \( i = 1, 2, \ldots, n \)
3. \( \lambda_1 + \lambda_2 + \ldots + \lambda_n = 1 \)

### Step 2: Apply the triangle inequality
Using the triangle inequality, we can express the modulus of the sum:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \leq | \lambda_1 a_1 | + | \lambda_2 a_2 | + \ldots + | \lambda_n a_n | \]

### Step 3: Simplify each term
Since \( |a_i| < 1 \), we can write:
\[ | \lambda_i a_i | = \lambda_i |a_i| < \lambda_i \]
Thus, we can substitute this back into our inequality:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | < \lambda_1 + \lambda_2 + \ldots + \lambda_n \]

### Step 4: Use the condition on \( \lambda \)
From the problem statement, we know that:
\[ \lambda_1 + \lambda_2 + \ldots + \lambda_n = 1 \]
Therefore, we can conclude:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | < 1 \]

### Step 5: Conclusion
Since the sum \( | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | \) is less than 1, we can conclude that:
\[ | \lambda_1 a_1 + \lambda_2 a_2 + \ldots + \lambda_n a_n | < 1 \]

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

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