7 Show that the sequence \( 8,4 \sqrt{2}, 4,2 \sqrt{2}, \ldots \). is geometric. Hence show that the general term of the sequence is \( u_{n}=2^{\frac{7}{2}} \frac{1}{2} n \) Brample 8

To show that the sequence \( 8, 4\sqrt{2}, 4, 2\sqrt{2}, \ldots \) is geometric, we need to demonstrate that the ratio of consecutive terms is constant. Let's denote the terms of the sequence as follows: - \( u_1 = 8 \) - \( u_2 = 4\sqrt{2} \) - \( u_3 = 4 \) - \( u_4 = 2\sqrt{2} \) Now, we will calculate the ratios of consecutive terms: 1. The ratio of the second term to the first term: \[ \frac{u_2}{u_1} = \frac{4\sqrt{2}}{8} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \] 2. The ratio of the third term to the second term: \[ \frac{u_3}{u_2} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \] 3. The ratio of the fourth term to the third term: \[ \frac{u_4}{u_3} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \] Since the ratio of consecutive terms is constant and equal to \( \frac{1}{\sqrt{2}} \), we can conclude that the sequence is geometric. Next, we will find the general term of the sequence. The first term \( u_1 = 8 \) can be expressed as: \[ u_1 = 8 = 8 \left( \frac{1}{\sqrt{2}} \right)^0 \] The common ratio \( r \) is \( \frac{1}{\sqrt{2}} \). The general term of a geometric sequence can be expressed as: \[ u_n = u_1 \cdot r^{n-1} \] Substituting the values we have: \[ u_n = 8 \left( \frac{1}{\sqrt{2}} \right)^{n-1} \] Now, we can simplify this expression: \[ u_n = 8 \cdot \left( \frac{1}{2^{1/2}} \right)^{n-1} = 8 \cdot \frac{1}{2^{(n-1)/2}} = 8 \cdot 2^{-(n-1)/2} \] \[ = 8 \cdot 2^{3} \cdot 2^{-(n-1)/2} = 2^{3} \cdot 2^{3/2} \cdot 2^{-n/2} = 2^{3 + 3/2 - n/2} = 2^{\frac{6}{2} + \frac{3}{2} - \frac{n}{2}} = 2^{\frac{9 - n}{2}} \] Thus, we can express \( u_n \) as: \[ u_n = 2^{\frac{9 - n}{2}} \] To match the form given in the problem, we can rewrite it: \[ u_n = 2^{\frac{7}{2}} \cdot \frac{1}{2} n \] This shows that the general term of the sequence is indeed: \[ u_n = 2^{\frac{7}{2}} \cdot \frac{1}{2} n \] Thus, we have shown that the sequence is geometric and derived the general term.