If `a_n>1` for all `n in N` then `log_(a_2) a_1+log_(a_3) a_2+.....log_(a_1)a_n` has the minimum value of

Options

A. 1
B. 2
C. 0
D. 3

Answer

To find the minimum value of the expression \( \log_{a_2} a_1 + \log_{a_3} a_2 + \ldots + \log_{a_n} a_1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic terms
We can use the change of base formula for logarithms, which states that \( \log_b a = \frac{\log a}{\log b} \). Therefore, we can rewrite each term in the sum:

\[ \log_{a_2} a_1 = \frac{\log a_1}{\log a_2}, \quad \log_{a_3} a_2 = \frac{\log a_2}{\log a_3}, \quad \ldots, \quad \log_{a_n} a_1 = \frac{\log a_1}{\log a_n} \]

Thus, the entire expression can be rewritten as:

\[ \frac{\log a_1}{\log a_2} + \frac{\log a_2}{\log a_3} + \ldots + \frac{\log a_n}{\log a_1} \]

### Step 2: Define the sum
Let \( S = \frac{\log a_1}{\log a_2} + \frac{\log a_2}{\log a_3} + \ldots + \frac{\log a_n}{\log a_1} \).

### Step 3: Apply the AM-GM Inequality
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms in \( S \):

\[ \frac{\frac{\log a_1}{\log a_2} + \frac{\log a_2}{\log a_3} + \ldots + \frac{\log a_n}{\log a_1}}{n} \geq \sqrt[n]{\frac{\log a_1}{\log a_2} \cdot \frac{\log a_2}{\log a_3} \cdots \frac{\log a_n}{\log a_1}} \]

### Step 4: Simplify the product
Notice that the product simplifies as follows:

\[ \frac{\log a_1}{\log a_2} \cdot \frac{\log a_2}{\log a_3} \cdots \frac{\log a_n}{\log a_1} = \frac{\log a_1}{\log a_1} = 1 \]

### Step 5: Conclude the inequality
Thus, we have:

\[ \frac{S}{n} \geq 1 \implies S \geq n \]

### Step 6: Determine the minimum value
The minimum value of \( S \) occurs when all the terms are equal, which happens when \( a_1 = a_2 = a_3 = \ldots = a_n \). Therefore, the minimum value of the expression is:

\[ \boxed{n} \]