Ты просишь вычислить первые пять членов последовательности по заданной формуле n-го члена для заданных последовательностей.

Давай вычислим первые пять членов последовательности для каждого случая. Для этого подставим $n = 1, 2, 3, 4, 5$ в каждую формулу. 1. a) $y_n = 3 - 2n$: - $y_1 = 3 - 2(1) = 1$ - $y_2 = 3 - 2(2) = -1$ - $y_3 = 3 - 2(3) = -3$ - $y_4 = 3 - 2(4) = -5$ - $y_5 = 3 - 2(5) = -7$ б) $y_n = 2n^2 - n$: - $y_1 = 2(1)^2 - 1 = 1$ - $y_2 = 2(2)^2 - 2 = 6$ - $y_3 = 2(3)^2 - 3 = 15$ - $y_4 = 2(4)^2 - 4 = 28$ - $y_5 = 2(5)^2 - 5 = 45$ в) $y_n = n^3 - 1$: - $y_1 = (1)^3 - 1 = 0$ - $y_2 = (2)^3 - 1 = 7$ - $y_3 = (3)^3 - 1 = 26$ - $y_4 = (4)^3 - 1 = 63$ - $y_5 = (5)^3 - 1 = 124$ г) $y_n = \frac{3n - 1}{2n}$: - $y_1 = \frac{3(1) - 1}{2(1)} = \frac{2}{2} = 1$ - $y_2 = \frac{3(2) - 1}{2(2)} = \frac{5}{4} = 1.25$ - $y_3 = \frac{3(3) - 1}{2(3)} = \frac{8}{6} = \frac{4}{3} \approx 1.33$ - $y_4 = \frac{3(4) - 1}{2(4)} = \frac{11}{8} = 1.375$ - $y_5 = \frac{3(5) - 1}{2(5)} = \frac{14}{10} = 1.4$ 2. a) $y_n = (-1)^n$: - $y_1 = (-1)^1 = -1$ - $y_2 = (-1)^2 = 1$ - $y_3 = (-1)^3 = -1$ - $y_4 = (-1)^4 = 1$ - $y_5 = (-1)^5 = -1$ б) $y_n = \frac{(-2)^n}{n^2 + 1}$: - $y_1 = \frac{(-2)^1}{1^2 + 1} = \frac{-2}{2} = -1$ - $y_2 = \frac{(-2)^2}{2^2 + 1} = \frac{4}{5} = 0.8$ - $y_3 = \frac{(-2)^3}{3^2 + 1} = \frac{-8}{10} = -0.8$ - $y_4 = \frac{(-2)^4}{4^2 + 1} = \frac{16}{17} \approx 0.94$ - $y_5 = \frac{(-2)^5}{5^2 + 1} = \frac{-32}{26} = -\frac{16}{13} \approx -1.23$ в) $y_n = (-1)^n \frac{1}{10^n}$: - $y_1 = (-1)^1 \frac{1}{10^1} = -\frac{1}{10} = -0.1$ - $y_2 = (-1)^2 \frac{1}{10^2} = \frac{1}{100} = 0.01$ - $y_3 = (-1)^3 \frac{1}{10^3} = -\frac{1}{1000} = -0.001$ - $y_4 = (-1)^4 \frac{1}{10^4} = \frac{1}{10000} = 0.0001$ - $y_5 = (-1)^5 \frac{1}{10^5} = -\frac{1}{100000} = -0.00001$ г) $y_n = \frac{(-1)^n + 2}{3n - 2}$: - $y_1 = \frac{(-1)^1 + 2}{3(1) - 2} = \frac{-1 + 2}{3 - 2} = \frac{1}{1} = 1$ - $y_2 = \frac{(-1)^2 + 2}{3(2) - 2} = \frac{1 + 2}{6 - 2} = \frac{3}{4} = 0.75$ - $y_3 = \frac{(-1)^3 + 2}{3(3) - 2} = \frac{-1 + 2}{9 - 2} = \frac{1}{7} \approx 0.14$ - $y_4 = \frac{(-1)^4 + 2}{3(4) - 2} = \frac{1 + 2}{12 - 2} = \frac{3}{10} = 0.3$ - $y_5 = \frac{(-1)^5 + 2}{3(5) - 2} = \frac{-1 + 2}{15 - 2} = \frac{1}{13} \approx 0.08$ 3. a) $y_n = 3 \cos(\frac{2\pi}{n})$: - $y_1 = 3 \cos(\frac{2\pi}{1}) = 3 \cos(2\pi) = 3(1) = 3$ - $y_2 = 3 \cos(\frac{2\pi}{2}) = 3 \cos(\pi) = 3(-1) = -3$ - $y_3 = 3 \cos(\frac{2\pi}{3}) = 3(-\frac{1}{2}) = -\frac{3}{2} = -1.5$ - $y_4 = 3 \cos(\frac{2\pi}{4}) = 3 \cos(\frac{\pi}{2}) = 3(0) = 0$ - $y_5 = 3 \cos(\frac{2\pi}{5}) \approx 3(0.309) \approx 0.927$ б) $y_n = \tan((-1)^n \frac{\pi}{4})$: - $y_1 = \tan((-1)^1 \frac{\pi}{4}) = \tan(-\frac{\pi}{4}) = -1$ - $y_2 = \tan((-1)^2 \frac{\pi}{4}) = \tan(\frac{\pi}{4}) = 1$ - $y_3 = \tan((-1)^3 \frac{\pi}{4}) = \tan(-\frac{\pi}{4}) = -1$ - $y_4 = \tan((-1)^4 \frac{\pi}{4}) = \tan(\frac{\pi}{4}) = 1$ - $y_5 = \tan((-1)^5 \frac{\pi}{4}) = \tan(-\frac{\pi}{4}) = -1$ в) $y_n = 1 - \cos^2(\frac{\pi}{n})$: - $y_1 = 1 - \cos^2(\frac{\pi}{1}) = 1 - \cos^2(\pi) = 1 - (-1)^2 = 1 - 1 = 0$ - $y_2 = 1 - \cos^2(\frac{\pi}{2}) = 1 - \cos^2(\frac{\pi}{2}) = 1 - (0)^2 = 1 - 0 = 1$ - $y_3 = 1 - \cos^2(\frac{\pi}{3}) = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$ - $y_4 = 1 - \cos^2(\frac{\pi}{4}) = 1 - (\frac{\sqrt{2}}{2})^2 = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2} = 0.5$ - $y_5 = 1 - \cos^2(\frac{\pi}{5}) \approx 1 - (0.809)^2 \approx 1 - 0.6545 \approx 0.3455$ г) $y_n = \sin(\pi n) - \cos(\pi n)$: - $y_1 = \sin(\pi (1)) - \cos(\pi (1)) = \sin(\pi) - \cos(\pi) = 0 - (-1) = 1$ - $y_2 = \sin(\pi (2)) - \cos(\pi (2)) = \sin(2\pi) - \cos(2\pi) = 0 - 1 = -1$ - $y_3 = \sin(\pi (3)) - \cos(\pi (3)) = \sin(3\pi) - \cos(3\pi) = 0 - (-1) = 1$ - $y_4 = \sin(\pi (4)) - \cos(\pi (4)) = \sin(4\pi) - \cos(4\pi) = 0 - 1 = -1$ - $y_5 = \sin(\pi (5)) - \cos(\pi (5)) = \sin(5\pi) - \cos(5\pi) = 0 - (-1) = 1$