ГДЗ по алгебре 9 класс Макарычев Задание 670

\[\boxed{\text{670}\text{\ (670)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ c_{n} = - 2n^{2} + 7\]

\[c_{1} = - 2 \cdot 1^{2} + 7 = 5\]

\[c_{2} = - 2 \cdot 2^{2} + 7 = - 1\]

\[c_{3} = - 2 \cdot 3^{2} + 7 = - 11\]

\[c_{4} = - 2 \cdot 4^{2} + 7 = - 25\]

\[c_{5} = - 2 \cdot 5^{2} + 7 = - 43.\]

\[\textbf{б)}\ c_{n} = \frac{100}{n² - 5}\]

\[c_{1} = \frac{100}{1^{2} - 5} = - 25\]

\[c_{2} = \frac{100}{2² - 5} = - 100\]

\[c_{3} = \frac{100}{3² - 5} = 25\]

\[c_{4} = \frac{100}{4² - 5} = \frac{100}{11} = 9\frac{1}{11}\]

\[c_{5} = \frac{100}{5² - 5} = 5.\]

\[\textbf{в)}\ c_{n} = - 2,5 \cdot 2^{n}\]

\[c_{1} = - 2,5 \cdot 2^{1} = - 5\]

\[c_{2} = - 2,5 \cdot 2^{2} = - 10\]

\[c_{3} = - 2,5 \cdot 2^{3} = - 20\]

\[c_{4} = - 2,5 \cdot 2^{4} = - 40\]

\[c_{5} = - 2,5 \cdot 2^{5} = - 80.\]

\[\textbf{г)}\ c_{n} = 3,2 \cdot 2^{- n}\]

\[c_{1} = 3,2 \cdot 2^{- 1} = 1,6\]

\[c_{2} = 3,2 \cdot 2^{- 2} = 0,8\]

\[c_{3} = 3,2 \cdot 2^{- 3} = 0,4\]

\[c_{4} = 3,2 \cdot 2^{- 4} = 0,2\]

\[c^{5} = 3,2 \cdot 2^{- 5} = 0,1.\]

\[\textbf{д)}\ c_{n} = \frac{( - 1)^{n - 1}}{4n}\]

\[c_{1} = \frac{( - 1)^{0}}{4} = \frac{1}{4}\]

\[c_{2} = \ \frac{( - 1)^{1}}{4 \cdot 2} = - \frac{1}{8}\]

\[c_{3} = \frac{( - 1)²}{4 \cdot 3} = \frac{1}{12}\]

\[c_{4} = \frac{( - 1)³}{4 \cdot 4} = - \frac{1}{16}\]

\[c_{5} = \frac{( - 1)^{4}}{4 \cdot 5} = \frac{1}{20}.\]

\[\textbf{е)}\ c_{n} = \frac{1 - ( - 1)^{n}}{2n + 1}\]

\[c_{1} = \frac{1 - ( - 1)}{2 + 1} = \frac{2}{3}\]

\[c_{2} = \ \frac{1 - ( - 1)²}{2 \cdot 2 + 1} = 0\]

\[c_{3} = \frac{1 - ( - 1)³}{2 \cdot 3 + 1} = \frac{2}{7}\]

\[c_{4} = \frac{1 - ( - 1)^{4}}{2 \cdot 4 + 1} = 0\]

\[c_{5} = \frac{1 - ( - 1)^{5}}{2 \cdot 5 + 1} = \frac{2}{11}.\]