Страница 140 номер 6.1.3, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Подставляем в формулу \(n\)-го члена номера 1, 2, 3 и 4.
1) \(a_n = -n + 10\): \(a_1 = -1 + 10 = 9\), \(a_2 = -2 + 10 = 8\), \(a_3 = -3 + 10 = 7\), \(a_4 = -4 + 10 = 6\).
2) \(a_n = 2n - 17\): \(a_1 = 2 \cdot 1 - 17 = -15\), \(a_2 = 2 \cdot 2 - 17 = -13\), \(a_3 = 2 \cdot 3 - 17 = -11\), \(a_4 = 2 \cdot 4 - 17 = -9\).
3) \(b_n = n^2 + n\): \(b_1 = 1^2 + 1 = 2\), \(b_2 = 2^2 + 2 = 6\), \(b_3 = 3^2 + 3 = 12\), \(b_4 = 4^2 + 4 = 20\).
4) \(b_n = n^2 - n\): \(b_1 = 1^2 - 1 = 0\), \(b_2 = 2^2 - 2 = 2\), \(b_3 = 3^2 - 3 = 6\), \(b_4 = 4^2 - 4 = 12\).
5) \(m_n = \frac{1}{n^3}\): \(m_1 = \frac{1}{1^3} = 1\), \(m_2 = \frac{1}{2^3} = \frac{1}{8}\), \(m_3 = \frac{1}{3^3} = \frac{1}{27}\), \(m_4 = \frac{1}{4^3} = \frac{1}{64}\).
6) \(m_n = \frac{1}{(n+1)^3}\): \(m_1 = \frac{1}{2^3} = \frac{1}{8}\), \(m_2 = \frac{1}{3^3} = \frac{1}{27}\), \(m_3 = \frac{1}{4^3} = \frac{1}{64}\), \(m_4 = \frac{1}{5^3} = \frac{1}{125}\).
7) \(c_n = \frac{n}{n+1}\): \(c_1 = \frac{1}{1+1} = \frac{1}{2}\), \(c_2 = \frac{2}{2+1} = \frac{2}{3}\), \(c_3 = \frac{3}{3+1} = \frac{3}{4}\), \(c_4 = \frac{4}{4+1} = \frac{4}{5}\).
8) \(c_n = \frac{n-1}{n+2}\): \(c_1 = \frac{1-1}{1+2} = 0\), \(c_2 = \frac{2-1}{2+2} = \frac{1}{4}\), \(c_3 = \frac{3-1}{3+2} = \frac{2}{5}\), \(c_4 = \frac{4-1}{4+2} = \frac{3}{6} = \frac{1}{2}\).
Ответ: 1) 9, 8, 7, 6; 2) \(-15\), \(-13\), \(-11\), \(-9\); 3) 2, 6, 12, 20; 4) 0, 2, 6, 12; 5) 1, \(\frac{1}{8}\), \(\frac{1}{27}\), \(\frac{1}{64}\); 6) \(\frac{1}{8}\), \(\frac{1}{27}\), \(\frac{1}{64}\), \(\frac{1}{125}\); 7) \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{4}{5}\); 8) 0, \(\frac{1}{4}\), \(\frac{2}{5}\), \(\frac{1}{2}\).
Первые четыре члена получаем, подставляя в формулу \(n\)-го члена номера 1, 2, 3 и 4.
1) \(a_n = -n + 10\): \(a_1 = -1 + 10 = 9\), \(a_2 = -2 + 10 = 8\), \(a_3 = -3 + 10 = 7\), \(a_4 = -4 + 10 = 6\).
2) \(a_n = 2n - 17\): \(a_1 = 2 \cdot 1 - 17 = -15\), \(a_2 = 2 \cdot 2 - 17 = -13\), \(a_3 = 2 \cdot 3 - 17 = -11\), \(a_4 = 2 \cdot 4 - 17 = -9\).
3) \(b_n = n^2 + n\): \(b_1 = 1^2 + 1 = 2\), \(b_2 = 2^2 + 2 = 6\), \(b_3 = 3^2 + 3 = 12\), \(b_4 = 4^2 + 4 = 20\).
4) \(b_n = n^2 - n\): \(b_1 = 1^2 - 1 = 0\), \(b_2 = 2^2 - 2 = 2\), \(b_3 = 3^2 - 3 = 6\), \(b_4 = 4^2 - 4 = 12\).
5) \(m_n = \frac{1}{n^3}\): \(m_1 = \frac{1}{1^3} = 1\), \(m_2 = \frac{1}{2^3} = \frac{1}{8}\), \(m_3 = \frac{1}{3^3} = \frac{1}{27}\), \(m_4 = \frac{1}{4^3} = \frac{1}{64}\).
6) \(m_n = \frac{1}{(n+1)^3}\): \(m_1 = \frac{1}{2^3} = \frac{1}{8}\), \(m_2 = \frac{1}{3^3} = \frac{1}{27}\), \(m_3 = \frac{1}{4^3} = \frac{1}{64}\), \(m_4 = \frac{1}{5^3} = \frac{1}{125}\).
7) \(c_n = \frac{n}{n+1}\): \(c_1 = \frac{1}{1+1} = \frac{1}{2}\), \(c_2 = \frac{2}{2+1} = \frac{2}{3}\), \(c_3 = \frac{3}{3+1} = \frac{3}{4}\), \(c_4 = \frac{4}{4+1} = \frac{4}{5}\).
8) \(c_n = \frac{n-1}{n+2}\): \(c_1 = \frac{1-1}{1+2} = 0\), \(c_2 = \frac{2-1}{2+2} = \frac{1}{4}\), \(c_3 = \frac{3-1}{3+2} = \frac{2}{5}\), \(c_4 = \frac{4-1}{4+2} = \frac{3}{6} = \frac{1}{2}\).
Ответ: 1) 9, 8, 7, 6; 2) \(-15\), \(-13\), \(-11\), \(-9\); 3) 2, 6, 12, 20; 4) 0, 2, 6, 12; 5) 1, \(\frac{1}{8}\), \(\frac{1}{27}\), \(\frac{1}{64}\); 6) \(\frac{1}{8}\), \(\frac{1}{27}\), \(\frac{1}{64}\), \(\frac{1}{125}\); 7) \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{4}{5}\); 8) 0, \(\frac{1}{4}\), \(\frac{2}{5}\), \(\frac{1}{2}\).
