If a_n 1 for every n in N,then the minimum | vedclass

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(C) Since $a_n > 1$ for all $n \in N$,all logarithmic terms are positive.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for $n$ pos

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$n$

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(C) Since $a_n > 1$ for all $n \in N$,all logarithmic terms are positive.Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for $n$ positive terms:$\frac{\log_{a_2} a_1 + \log_{a_3} a_2 + \dots + \log_{a_1} a_n}{n} \ge \sqrt[n]{(\log_{a_2} a_1) \cdot (\log_{a_3} a_2) \cdot \dots \cdot (\log_{a_1} a_n)}$Using the change of base formula $\log_b a = \frac{\ln a}{\ln b}$,the product becomes:$\frac{\ln a_1}{\ln a_2} \cdot \frac{\ln a_2}{\ln a_3} \cdot \dots \cdot \frac{\ln a_n}{\ln a_1} = 1$Therefore,$\frac{\sum \log_{a_{i+1}} a_i}{n} \ge \sqrt[n]{1} = 1$$\sum \log_{a_{i+1}} a_i \ge n$Thus,the minimum value is $n$.

If $a_n > 1$ for every $n \in N$,then the minimum value of $\log_{a_2} a_1 + \log_{a_3} a_2 + \dots + \log_{a_n} a_{n-1} + \log_{a_1} a_n$ is:

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