Let I(n) = n^n and J(n) = 1 times 3 times 5 t | vedclass

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(A) We are given $I(n) = n^n$ and $J(n) = 1 	imes 3 	imes 5 	imes \ldots 	imes (2n - 1)$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ i

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$I(n) > J(n)$

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(A) We are given $I(n) = n^n$ and $J(n) = 1 	imes 3 	imes 5 	imes \ldots 	imes (2n - 1)$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the $n$ positive integers $1, 3, 5, \ldots, (2n - 1)$:$\frac{1 + 3 + 5 + \ldots + (2n - 1)}{n} > \sqrt[n]{1 	imes 3 	imes 5 	imes \ldots 	imes (2n - 1)}$The sum of the first $n$ odd integers is $n^2$,so $\frac{n^2}{n} > (J(n))^{1/n}$.This simplifies to $n > (J(n))^{1/n}$.Raising both sides to the power of $n$,we get $n^n > J(n)$.Thus,$I(n) > J(n)$.

Let $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$ for all $n > 1, n \in N$. Then:

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