The geometric mean of the sequence 1, 2, 2^2, | vedclass

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(A) The sequence is $1, 2, 2^2, ..., 2^n$. The total number of terms is $n+1$.The geometric mean $GM$ is given by the $(n+1)$-th root of the product o

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$2^{\frac{n}{2}}$

Vedclass · Answer

(A) The sequence is $1, 2, 2^2, ..., 2^n$. The total number of terms is $n+1$.The geometric mean $GM$ is given by the $(n+1)$-th root of the product of the terms:$GM = (1 	imes 2 	imes 2^2 	imes ... 	imes 2^n)^{\frac{1}{n+1}}$Using the property of exponents,the product is:$1 	imes 2^1 	imes 2^2 	imes ... 	imes 2^n = 2^{0+1+2+...+n}$The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$,so:$2^{0+1+2+...+n} = 2^{\frac{n(n+1)}{2}}$Now,substitute this back into the $GM$ formula:$GM = (2^{\frac{n(n+1)}{2}})^{\frac{1}{n+1}} = 2^{\frac{n(n+1)}{2(n+1)}} = 2^{\frac{n}{2}}$

The geometric mean of the sequence $1, 2, 2^2, ..., 2^n$ is:

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