If an > 1 for all n in N , then 1/n [log...

If `a_n > 1` for all `n in N` , then `1/n [log_(a_2) a_1 + log a_3a_2 + ….+ log_(a_n) a_(n-1) + log_(a_1) a_n]` has the least value :

A

`e/2`

B

`(3e)/2`

C

`(5e)/2`

D

`(7e)/2`

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The correct Answer is:

B

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