91Ó°ÊÓ

We are given the hint to rewrite the expression for \(N\). Using the hint, we can write: $$\frac{N}{2^{n}}=\frac{(2 n) !}{2^{2 n}(n !)^{2}}=\frac{1.3.5 \ldots(2 n-1)}{2.4.6 \ldots .(2 n)} \cdot \frac{2.4.6 \ldots . (2 n)}{2.4.6 \ldots(2 n)}$$

To show the left side of the inequality \(\frac{2^{2 n}}{2 \sqrt{n}}(2 n+1) \cdot \frac{N^{2}}{2^{4 n}}>2 n \cdot \frac{N^{2}}{2^{4 n}}$$ Taking only the first half of this inequality, we can write this as a product of fractions: $$\frac{1}{1}>\frac{1}{2}\frac{3}{4}\frac{5}{6}\ldots \frac{2n-1}{2n}$$ Notice that every fraction in this product is less than 1: $$\frac{k}{k+1}<1, \text { for } k=1,3,5,\ldots,2n-1$$ Now, we can multiply both sides of the inequality by \(2^n\), and we get: $$\frac{2^n}{2^n}>\frac{1}{2^n}\frac{3}{2^n}\frac{5}{2^n}\ldots \frac{2n-1}{2^n}\frac{2^n}{2^n}$$ Now, using the expression obtained in step 1, which is \(\frac{N}{2^n}=\frac{1\cdot3\cdot5\ldots(2n-1)}{2^n}\), we can replace the right side with \(N\) $$2^n>N$$ Squaring both sides, we get: $$2^{2n}>N^2$$ Dividing both sides by \(2\sqrt{n}\), we have: $$\frac{2^{2n}}{2\sqrt{n}}<N$$

Now, we will show the right side of the inequality \(N<\frac{2^{2 n}}{\sqrt{2 n}}\). We can use similar techniques as we did for the left side of the inequality. We start with the expression we derived earlier: $$\frac{N}{2^n}=\frac{1\cdot3\cdot5\ldots(2n-1)}{2^n}$$ Now, we can write a product of fractions with each fraction greater than 1: $$\frac{1}{1}<\frac{2}{1}\frac{4}{3}\frac{6}{5}\ldots \frac{2n}{2n-1}$$ Notice that every fraction in this product is greater than 1: $$\frac{k+1}{k}>1, \text { for } k=1,3,5,\ldots,2n-1$$ We multiply both sides by \(2^n\), and we get: $$\frac{2^n}{2^n}<\frac{2^{n+1}}{2^n}\frac{2^{n+2}}{2^n}\frac{2^{n+4}}{2^n}\ldots \frac{2^{n(2n)}}{2^n}$$ This simplifies to: $$2^n<N<2^{2n}$$ Now divide both sides by \(\sqrt{2 n}\): $$N<\frac{2^{2n}}{\sqrt{2 n}}$$ Hence, we have shown both the left and right sides of the inequality: $$\frac{2^{2 n}}{2 \sqrt{n}}<N<\frac{2^{2 n}}{\sqrt{2 n}}$$