Szczegółowe wyjaśnienie:
Podstawiamy za n kolejne liczby {1, 2, 3, 4, 5, 6}.
[tex]a)\ a_n=\dfrac{n^2}{n^3+1}\\\\a_1=\dfrac{1^2}{1^3+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\\\\a_2=\dfrac{2^2}{2^3+1}=\dfrac{4}{8+1}=\dfrac{4}{9}\\\\a_3=\dfrac{3^2}{3^3+1}=\dfrac{9}{27+1}=\dfrac{9}{28}\\\\a_4=\dfrac{4^2}{4^3+1}=\dfrac{16}{64+1}=\dfrac{16}{65}\\\\a_5=\dfrac{5^2}{5^3+1}=\dfrac{25}{125+1}=\dfrac{25}{126}\\\\a_6=\dfrac{6^2}{6^3+1}=\dfrac{36}{216+1}=\dfrac{35}{217}[/tex]
[tex]b)\ a_n=n^2-3n+1\\\\a_1=1^2-3\cdot1+1=1-3+1=-1\\\\a_2=2^2-3\cdot2+1=4-6+1=-1\\\\a_3=3^2-3\cdot3+1=9-9+1=1\\\\a_4=4^2-3\cdot4+1=16-12+1=5\\\\a_5=5^2-3\cdot5+1=25-15+1=11\\\\a_6=6^2-3\cdot6+1=36-18+1=19[/tex]
[tex]c)\ a_n=3n^2-7n-11\\\\a_1=3\cdot1^2-7\cdot1-11=3-7-11=-15\\\\a_2=3\cdot2^2-7\cdot2-11=14-14-11=-11\\\\a_3=3\cdot3^2-7\cdot3-11=27-21-11=-5\\\\a_4=3\cdot4^2-7\cdot4-11=48-28-11=9\\\\a_5=3\cdot5^2-7\cdot5-11=75-35-11=29\\\\a_6=3\cdot6^2-7\cdot6-11=108-56-11=41[/tex]
