If `a_(i)gt0` for `i u=1, 2, 3, … ,n` and `a_(1)a_(2)…a_(n)=1,` then the minimum value of `(1+a_(1))(1+a_(2))…(1+a_(n))`, is

Options

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

Explanation

To find the minimum value of the expression \((1 + a_1)(1 + a_2) \cdots (1 + a_n)\) given that \(a_i > 0\) for \(i = 1, 2, \ldots, n\) and \(a_1 a_2 \cdots a_n = 1\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for any positive numbers \(x_1, x_2, \ldots, x_n\):
\[ \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} \]
We will apply this to the terms \(1 + a_i\).

2. **Set Up the Inequality**:
For each \(i\), we can write:
\[ 1 + a_i \geq 2\sqrt{1 \cdot a_i} \]
This can be derived from the AM-GM inequality applied to the numbers \(1\) and \(a_i\).

3. **Multiply the Inequalities**:
Now, we can multiply these inequalities for all \(i\) from \(1\) to \(n\):
\[ (1 + a_1)(1 + a_2) \cdots (1 + a_n \geq (2\sqrt{1 \cdot a_1})(2\sqrt{1 \cdot a_2}) \cdots (2\sqrt{1 \cdot a_n}) \]
This simplifies to:
\[ (1 + a_1)(1 + a_2) \cdots (1 + a_n \geq 2^n \sqrt{a_1 a_2 \cdots a_n} \]

4. **Substitute the Given Condition**:
Since we know that \(a_1 a_2 \cdots a_n = 1\), we can substitute this into our inequality:
\[ (1 + a_1)(1 + a_2) \cdots (1 + a_n \geq 2^n \sqrt{1} = 2^n \]

5. **Conclusion**:
Therefore, the minimum value of \((1 + a_1)(1 + a_2) \cdots (1 + a_n)\) is \(2^n\).

### Final Answer:
The minimum value of \((1 + a_1)(1 + a_2) \cdots (1 + a_n)\) is \(2^n\).