Страница 91 номер 230, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением
Записать первые шесть членов последовательности, заданной условием \(a_1 = 1\) и рекуррентной формулой: 1) \(a_{n+1} = \sqrt{a_n^2 + 3}\); 2) \(a_{n+1} = \sqrt{\frac{a_n^2}{3}}\).
1) \(a_{n + 1} = \sqrt{a_{n}^{2} + 3}\)
\(a_{1} = 1\)
\(a_{2} = \sqrt{a_{1}^{2} + 3} = \sqrt{1^{2} + 3} = \sqrt{4} = 2\)
\(a_{3} = \sqrt{a_{2}^{2} + 3} = \sqrt{2^{2} + 3} = \sqrt{4 + 3} = \sqrt{7}\)
\(a_{4} = \sqrt{a_{3}^{2} + 3} = \sqrt{\left(\sqrt{7}\right)^{2} + 3} = \sqrt{7 + 3} = \sqrt{10}\)
\(a_{5} = \sqrt{a_{4}^{2} + 3} = \sqrt{\left(\sqrt{10}\right)^{2} + 3} = \sqrt{10 + 3} = \sqrt{13}\)
\(a_{6} = \sqrt{a_{5}^{2} + 3} = \sqrt{\left(\sqrt{13}\right)^{2} + 3} = \sqrt{13 + 3} = \sqrt{16} = 4\)
Ответ: \(1\), \(2\), \(\sqrt{7}\), \(\sqrt{10}\), \(\sqrt{13}\), \(4\)
2) \(a_{n + 1} = \sqrt{\frac{a_{n}^{2}}{3}}\)
\(a_{1} = 1\)
\(a_{2} = \sqrt{\dfrac{a_{1}^{2}}{3}} = \sqrt{\dfrac{1}{3}} = \dfrac{1}{\sqrt{3}}\)
\(a_{3} = \sqrt{\dfrac{a_{2}^{2}}{3}} = \sqrt{\dfrac{\left(\frac{1}{\sqrt{3}}\right)^{2}}{3}} = \sqrt{\dfrac{\frac{1}{3}}{3}} = \sqrt{\dfrac{1}{3} \cdot \dfrac{1}{3}} = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3}\)
\(a_{4} = \sqrt{\dfrac{a_{3}^{2}}{3}} = \sqrt{\dfrac{\left(\frac{1}{3}\right)^{2}}{3}} = \sqrt{\dfrac{\frac{1}{9}}{3}} = \sqrt{\dfrac{1}{9} \cdot \dfrac{1}{3}} = \sqrt{\dfrac{1}{27}} = \dfrac{1}{\sqrt{27}} = \dfrac{1}{3\sqrt{3}}\)
\(a_{5} = \sqrt{\dfrac{a_{4}^{2}}{3}} = \sqrt{\dfrac{\left(\frac{1}{3\sqrt{3}}\right)^{2}}{3}} = \sqrt{\dfrac{\frac{1}{9 \cdot 3}}{3}} = \sqrt{\dfrac{1}{27} \cdot \dfrac{1}{3}} = \sqrt{\dfrac{1}{81}} = \dfrac{1}{9}\)
\(a_{6} = \sqrt{\dfrac{a_{5}^{2}}{3}} = \sqrt{\dfrac{\left(\frac{1}{9}\right)^{2}}{3}} = \sqrt{\dfrac{\frac{1}{81}}{3}} =\)
\(= \sqrt{\dfrac{1}{81} \cdot \dfrac{1}{3}} = \sqrt{\dfrac{1}{81 \cdot 3}} = \dfrac{1}{\sqrt{81 \cdot 3}} = \dfrac{1}{9\sqrt{3}}\)
Ответ: \(1\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{3}\), \(\frac{1}{3\sqrt{3}}\), \(\frac{1}{9}\), \(\frac{1}{9\sqrt{3}}\)
