Do każdego wzoru podanego ciągu za n podstawiamy kolejno: 1, 2, 3, 4, 5, czyli kolejne wyrazy ciągu.
[tex]a) \ a_{n} = \frac{n-1}{n^{2}}\\\\a_1 = \frac{1-1}{1^{2}} = \frac{0}{1} = 0\\\\a_2 = \frac{2-1}{2^{2}} = \frac{1}{4}\\\\a_3 = \frac{3-1}{3^{2}} = \frac{2}{9}\\\\a_4 = \frac{4-1}{4^{2}} = \frac{3}{16}\\\\a_5 = \frac{5-1}{5^{2}} = \frac{4}{25}[/tex]
[tex]b) \ a_{n} =\frac{3n}{2n+4}\\\\a_1 = \frac{3\cdot1}{2\cdot1+4} = \frac{3}{2+4} = \frac{3}{6} =\frac{1}{2}\\\\a_2 = \frac{3\cdot2}{2\cdot2+4} = \frac{6}{4+4} = \frac{6}{8} = \frac{3}{4}\\\\a_3 = \frac{3\cdot3}{2\cdot3+4} = \frac{9}{6+4} = \frac{9}{10}\\\\a_4 = \frac{3\cdot4}{2\cdot4+4} = \frac{12}{8+4} = \frac{12}{12} = 1\\\\a_5 = \frac{3\cdot5}{2\cdot5+4} = \frac{15}{10+4} = \frac{15}{14} = 1\frac{1}{14}[/tex]
[tex]c) \ a_{n} = (-1)^{n+1}\cdot2^{n}\\\\a_1 = (-1)^{1+1}\cdot2^{1} = (-1)^{2}\cdot2 = 1\cdot2 = 2\\\\a_2 = (-1)^{2+1}\cdot2^{2} = (-1)^{3}\cdot4 = -1\cdot4 = -4\\\\a_3 = (-1)^{3+1}\cdot2^{3} = (-1)^{4}\cdot8=1\cdot8 = 8\\\\a_{4} = (-1)^{4+1}\cdot2^{4} = (-1)^{5}\cdot16 = -1\cdot16 = -16\\\\a_{5} = (-1)^{5+1}\cdot2^{5} = (-1)^{6}\cdot32 = 1\cdot32 = 32[/tex]
Wyjaśnienie
Liczba ujemna podniesiona do potęgi nieparzystej daje liczbę ujemną.
Liczba ujemna podniesiona do potęgi parzystej daje liczbę dodatnią.