If n is a natural number,then left( frac{n+1} | vedclass

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(B) We test the inequality $\left( \frac{n+1}{2} \right)^n \ge n!$ for different values of $n \in \mathbb{N}$.For $n = 1$: $\left( \frac{1+1}{2} \righ

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$n \ge 1$

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(B) We test the inequality $\left( \frac{n+1}{2} \right)^n \ge n!$ for different values of $n \in \mathbb{N}$.For $n = 1$: $\left( \frac{1+1}{2} \right)^1 = 1^1 = 1$ and $1! = 1$. Since $1 \ge 1$,the statement is true for $n = 1$.For $n = 2$: $\left( \frac{2+1}{2} \right)^2 = (1.5)^2 = 2.25$ and $2! = 2$. Since $2.25 \ge 2$,the statement is true for $n = 2$.For $n = 3$: $\left( \frac{3+1}{2} \right)^3 = 2^3 = 8$ and $3! = 6$. Since $8 \ge 6$,the statement is true for $n = 3$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for $n$ positive real numbers $x_1, x_2, \dots, x_n$,we have $\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$.Let the numbers be $1, 2, 3, \dots, n$. Then $\frac{1+2+3+\dots+n}{n} \ge \sqrt[n]{1 \cdot 2 \cdot 3 \dots n}$.Since $1+2+3+\dots+n = \frac{n(n+1)}{2}$,we get $\frac{n(n+1)}{2n} \ge \sqrt[n]{n!}$,which simplifies to $\frac{n+1}{2} \ge (n!)^{1/n}$.Raising both sides to the power $n$,we get $\left( \frac{n+1}{2} \right)^n \ge n!$.This inequality holds for all $n \ge 1$.

If $n$ is a natural number,then $\left( \frac{n+1}{2} \right)^n \ge n!$ is true when

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