По заданной формуле п-го члена вычислите первые пять членов последовательности (у ): 3. a) yn = sin - ctg (2n + 1); 6) Yn = cos + tg(2n + 1); B) yn = n sin + n2 cos; r) yn = sin-ncos.

Решим каждое задание по порядку, вычислив первые пять членов последовательности.

3. a) $$y_n = \sin{\frac{n\pi}{2}} - \operatorname{ctg}{\frac{\pi}{4}(2n + 1)}$$

1. n = 1: $$y_1 = \sin{\frac{\pi}{2}} - \operatorname{ctg}{\frac{3\pi}{4}} = 1 - (-1) = 2$$
2. n = 2: $$y_2 = \sin{\pi} - \operatorname{ctg}{\frac{5\pi}{4}} = 0 - 1 = -1$$
3. n = 3: $$y_3 = \sin{\frac{3\pi}{2}} - \operatorname{ctg}{\frac{7\pi}{4}} = -1 - (-1) = 0$$
4. n = 4: $$y_4 = \sin{2\pi} - \operatorname{ctg}{\frac{9\pi}{4}} = 0 - 1 = -1$$
5. n = 5: $$y_5 = \sin{\frac{5\pi}{2}} - \operatorname{ctg}{\frac{11\pi}{4}} = 1 - (-1) = 2$$

Ответ: 2, -1, 0, -1, 2

6) $$y_n = \cos{\frac{n\pi}{2}} + \operatorname{tg}{\frac{\pi}{4}(2n + 1)}$$

1. n = 1: $$y_1 = \cos{\frac{\pi}{2}} + \operatorname{tg}{\frac{3\pi}{4}} = 0 + (-1) = -1$$
2. n = 2: $$y_2 = \cos{\pi} + \operatorname{tg}{\frac{5\pi}{4}} = -1 + 1 = 0$$
3. n = 3: $$y_3 = \cos{\frac{3\pi}{2}} + \operatorname{tg}{\frac{7\pi}{4}} = 0 + (-1) = -1$$
4. n = 4: $$y_4 = \cos{2\pi} + \operatorname{tg}{\frac{9\pi}{4}} = 1 + 1 = 2$$
5. n = 5: $$y_5 = \cos{\frac{5\pi}{2}} + \operatorname{tg}{\frac{11\pi}{4}} = 0 + (-1) = -1$$

Ответ: -1, 0, -1, 2, -1

B) $$y_n = n \sin{\frac{n\pi}{2}} + n^2 \cos{\frac{n\pi}{2}}$$

1. n = 1: $$y_1 = 1 \cdot \sin{\frac{\pi}{2}} + 1^2 \cdot \cos{\frac{\pi}{2}} = 1 \cdot 1 + 1 \cdot 0 = 1$$
2. n = 2: $$y_2 = 2 \cdot \sin{\pi} + 2^2 \cdot \cos{\pi} = 2 \cdot 0 + 4 \cdot (-1) = -4$$
3. n = 3: $$y_3 = 3 \cdot \sin{\frac{3\pi}{2}} + 3^2 \cdot \cos{\frac{3\pi}{2}} = 3 \cdot (-1) + 9 \cdot 0 = -3$$
4. n = 4: $$y_4 = 4 \cdot \sin{2\pi} + 4^2 \cdot \cos{2\pi} = 4 \cdot 0 + 16 \cdot 1 = 16$$
5. n = 5: $$y_5 = 5 \cdot \sin{\frac{5\pi}{2}} + 5^2 \cdot \cos{\frac{5\pi}{2}} = 5 \cdot 1 + 25 \cdot 0 = 5$$

Ответ: 1, -4, -3, 16, 5

r) $$y_n = \sin{\frac{n\pi}{4}} - n \cos{\frac{n\pi}{4}}$$

1. n = 1: $$y_1 = \sin{\frac{\pi}{4}} - 1 \cdot \cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$$
2. n = 2: $$y_2 = \sin{\frac{\pi}{2}} - 2 \cdot \cos{\frac{\pi}{2}} = 1 - 2 \cdot 0 = 1$$
3. n = 3: $$y_3 = \sin{\frac{3\pi}{4}} - 3 \cdot \cos{\frac{3\pi}{4}} = \frac{\sqrt{2}}{2} - 3 \cdot (-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} = 2\sqrt{2}$$
4. n = 4: $$y_4 = \sin{\pi} - 4 \cdot \cos{\pi} = 0 - 4 \cdot (-1) = 4$$
5. n = 5: $$y_5 = \sin{\frac{5\pi}{4}} - 5 \cdot \cos{\frac{5\pi}{4}} = -\frac{\sqrt{2}}{2} - 5 \cdot (-\frac{\sqrt{2}}{2}) = -\frac{\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} = 2\sqrt{2}$$

Ответ: 0, 1, $$2\sqrt{2}$$, 4, $$2\sqrt{2}$$