Страница 208 номер 694, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите четыре первых члена последовательности \((a_n)\), заданной формулой \(n\)-го члена: 1) \(a_n = n + 4\); 2) \(a_n = 4n - 3\); 3) \(a_n = \frac{n}{n^2 + 1}\); 4) \(a_n = \frac{2^n}{n}\).
1) \(a_{n} = n + 4\)
\(a_{1} = 1 + 4 = 5;\)
\(a_{2} = 2 + 4 = 6\)
\(a_{3} = 3 + 4 = 7\)
\(a_{4} = 4 + 4 = 8\)
2) \(a_{n} = 4n - 3\)
\(a_{1} = 4 \cdot 1 - 3 = 1\)
\(a_{2} = 4 \cdot 2 - 3 = 5\)
\(a_{3} = 4 \cdot 3 - 3 = 9\)
\(a_{4} = 4 \cdot 4 - 3 = 13\)
3) \(a_{n} = \frac{n}{n^{2} + 1}\)
\(a_{1} = \dfrac{1}{1^{2} + 1} = \dfrac{1}{2}\)
\(a_{2} = \dfrac{2}{2^{2} + 1} = \dfrac{2}{5}\)
\(a_{3} = \dfrac{3}{3^{2} + 1} = \dfrac{3}{10}\)
\(a_{4} = \dfrac{4}{4^{2} + 1} = \dfrac{4}{17}\)
4) \(a_{n} = \frac{2^{n}}{n}\)
\(a_{1} = \dfrac{2^{1}}{1} = 2\)
\(a_{2} = \dfrac{2^{2}}{2} = 2\)
\(a_{3} = \dfrac{2^{3}}{3} = \dfrac{8}{3}\)
\(a_{4} = \dfrac{2^{4}}{4} = \dfrac{16}{4} = 4\)
Ответ: 1) \(5;\ 6;\ 7;\ 8;\ldots\); 2) \(1;\ 5;\ 9;\ 13;\ldots\); 3) \(\frac{1}{2};\frac{2}{5};\frac{3}{10};\frac{4}{17};\ldots\); 4) \(2;\ 2;\frac{8}{3};\ 4;\ldots\).
