Страница 140 номер 6.1.4, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Подставляем в формулу \(n\)-го члена номера 1, 6, 11 и 18.
1) \(a_n = 2n + 5\): \(a_1 = 2 \cdot 1 + 5 = 7\), \(a_6 = 2 \cdot 6 + 5 = 17\), \(a_{11} = 2 \cdot 11 + 5 = 27\), \(a_{18} = 2 \cdot 18 + 5 = 41\).
2) \(a_n = 2n - 4\): \(a_1 = 2 \cdot 1 - 4 = -2\), \(a_6 = 2 \cdot 6 - 4 = 8\), \(a_{11} = 2 \cdot 11 - 4 = 18\), \(a_{18} = 2 \cdot 18 - 4 = 32\).
3) \(b_n = -2n - 7\): \(b_1 = -2 \cdot 1 - 7 = -9\), \(b_6 = -2 \cdot 6 - 7 = -19\), \(b_{11} = -2 \cdot 11 - 7 = -29\), \(b_{18} = -2 \cdot 18 - 7 = -43\).
4) \(b_n = 1 - 3n\): \(b_1 = 1 - 3 \cdot 1 = -2\), \(b_6 = 1 - 3 \cdot 6 = -17\), \(b_{11} = 1 - 3 \cdot 11 = -32\), \(b_{18} = 1 - 3 \cdot 18 = -53\).
5) \(c_n = -n^2\): \(c_1 = -1^2 = -1\), \(c_6 = -6^2 = -36\), \(c_{11} = -11^2 = -121\), \(c_{18} = -18^2 = -324\).
6) \(c_n = -n^2 + 1\): \(c_1 = -1^2 + 1 = 0\), \(c_6 = -6^2 + 1 = -35\), \(c_{11} = -11^2 + 1 = -120\), \(c_{18} = -18^2 + 1 = -323\).
7) \(a_n = \frac{(-1)^n}{n+1}\); множитель \((-1)^n\) равен \(-1\) при нечётном номере и 1 при чётном, поэтому знаки чередуются: \(a_1 = \frac{(-1)^1}{1+1} = -\frac{1}{2}\), \(a_6 = \frac{(-1)^6}{6+1} = \frac{1}{7}\), \(a_{11} = \frac{(-1)^{11}}{11+1} = -\frac{1}{12}\), \(a_{18} = \frac{(-1)^{18}}{18+1} = \frac{1}{19}\).
8) \(a_n = \frac{(-1)^n}{n+3}\): \(a_1 = \frac{(-1)^1}{1+3} = -\frac{1}{4}\), \(a_6 = \frac{(-1)^6}{6+3} = \frac{1}{9}\), \(a_{11} = \frac{(-1)^{11}}{11+3} = -\frac{1}{14}\), \(a_{18} = \frac{(-1)^{18}}{18+3} = \frac{1}{21}\).
Ответ: 1) \(a_1 = 7\), \(a_6 = 17\), \(a_{11} = 27\), \(a_{18} = 41\); 2) \(a_1 = -2\), \(a_6 = 8\), \(a_{11} = 18\), \(a_{18} = 32\); 3) \(b_1 = -9\), \(b_6 = -19\), \(b_{11} = -29\), \(b_{18} = -43\); 4) \(b_1 = -2\), \(b_6 = -17\), \(b_{11} = -32\), \(b_{18} = -53\); 5) \(c_1 = -1\), \(c_6 = -36\), \(c_{11} = -121\), \(c_{18} = -324\); 6) \(c_1 = 0\), \(c_6 = -35\), \(c_{11} = -120\), \(c_{18} = -323\); 7) \(a_1 = -\frac{1}{2}\), \(a_6 = \frac{1}{7}\), \(a_{11} = -\frac{1}{12}\), \(a_{18} = \frac{1}{19}\); 8) \(a_1 = -\frac{1}{4}\), \(a_6 = \frac{1}{9}\), \(a_{11} = -\frac{1}{14}\), \(a_{18} = \frac{1}{21}\).
Нужные члены получаем, подставляя в формулу \(n\)-го члена номера 1, 6, 11 и 18.
1) \(a_n = 2n + 5\): \(a_1 = 2 \cdot 1 + 5 = 7\), \(a_6 = 2 \cdot 6 + 5 = 17\), \(a_{11} = 2 \cdot 11 + 5 = 27\), \(a_{18} = 2 \cdot 18 + 5 = 41\).
2) \(a_n = 2n - 4\): \(a_1 = 2 \cdot 1 - 4 = -2\), \(a_6 = 2 \cdot 6 - 4 = 8\), \(a_{11} = 2 \cdot 11 - 4 = 18\), \(a_{18} = 2 \cdot 18 - 4 = 32\).
3) \(b_n = -2n - 7\): \(b_1 = -2 \cdot 1 - 7 = -9\), \(b_6 = -2 \cdot 6 - 7 = -19\), \(b_{11} = -2 \cdot 11 - 7 = -29\), \(b_{18} = -2 \cdot 18 - 7 = -43\).
4) \(b_n = 1 - 3n\): \(b_1 = 1 - 3 \cdot 1 = -2\), \(b_6 = 1 - 3 \cdot 6 = -17\), \(b_{11} = 1 - 3 \cdot 11 = -32\), \(b_{18} = 1 - 3 \cdot 18 = -53\).
5) \(c_n = -n^2\): \(c_1 = -1^2 = -1\), \(c_6 = -6^2 = -36\), \(c_{11} = -11^2 = -121\), \(c_{18} = -18^2 = -324\).
6) \(c_n = -n^2 + 1\): \(c_1 = -1^2 + 1 = 0\), \(c_6 = -6^2 + 1 = -35\), \(c_{11} = -11^2 + 1 = -120\), \(c_{18} = -18^2 + 1 = -323\).
7) \(a_n = \frac{(-1)^n}{n+1}\). Множитель \((-1)^n\) равен \(-1\) при нечётном номере и 1 при чётном, поэтому знаки членов чередуются:
\(a_1 = \frac{(-1)^1}{1+1} = -\frac{1}{2}\), \(a_6 = \frac{(-1)^6}{6+1} = \frac{1}{7}\), \(a_{11} = \frac{(-1)^{11}}{11+1} = -\frac{1}{12}\), \(a_{18} = \frac{(-1)^{18}}{18+1} = \frac{1}{19}\).
8) \(a_n = \frac{(-1)^n}{n+3}\): \(a_1 = \frac{(-1)^1}{1+3} = -\frac{1}{4}\), \(a_6 = \frac{(-1)^6}{6+3} = \frac{1}{9}\), \(a_{11} = \frac{(-1)^{11}}{11+3} = -\frac{1}{14}\), \(a_{18} = \frac{(-1)^{18}}{18+3} = \frac{1}{21}\).
Ответ: 1) \(a_1 = 7\), \(a_6 = 17\), \(a_{11} = 27\), \(a_{18} = 41\); 2) \(a_1 = -2\), \(a_6 = 8\), \(a_{11} = 18\), \(a_{18} = 32\); 3) \(b_1 = -9\), \(b_6 = -19\), \(b_{11} = -29\), \(b_{18} = -43\); 4) \(b_1 = -2\), \(b_6 = -17\), \(b_{11} = -32\), \(b_{18} = -53\); 5) \(c_1 = -1\), \(c_6 = -36\), \(c_{11} = -121\), \(c_{18} = -324\); 6) \(c_1 = 0\), \(c_6 = -35\), \(c_{11} = -120\), \(c_{18} = -323\); 7) \(a_1 = -\frac{1}{2}\), \(a_6 = \frac{1}{7}\), \(a_{11} = -\frac{1}{12}\), \(a_{18} = \frac{1}{19}\); 8) \(a_1 = -\frac{1}{4}\), \(a_6 = \frac{1}{9}\), \(a_{11} = -\frac{1}{14}\), \(a_{18} = \frac{1}{21}\).
