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Text Solution

`2^(n)``2^((n)/(2))``2^(1//2)``2^(n-1)`

Answer :

BSolution :

Geometric mean `= root(n+1)(1.2.4.8…2^(n))` <br> `(because" there are "(n+1)" multiples from "2^(@)" to "2^(n))` <br> `=root(n+1)(2^(0).2^(1).2^(2).2^(3)…2^(n))` <br> `=root(n+1)(2^(1+2+3+…+n))=root(n+1)(2^((n(n+1))/(2)))` <br> `=[2^((n(n+1))/(2))]^((1)/(n+1))=2^((n)/(2))`