1. a) [latex]a_n=frac{5n+2}{15n-4}\a_1=frac{5+2}{15-4}=frac{7}{11}\a_2=frac{10+2}{30-4}=frac{12}{26}=frac{6}{13}\a_3=frac{15+2}{45-4}=frac{17}{41}\a_4=frac{20+2}{60-4}=frac{22}{56}=frac{11}{28}\a_5=frac{25+2}{75-4}=frac{27}{71}[/latex] b) [latex]a_n=n^{n-4}\a_1=1^{1-4}=1^[-3}=1\a_2=2^{2-4}=2^{-2}=frac{1}{4}\a_3=3^{3-4}=3^{-1}=frac{1}{3}\a_4=4^{4-4}=4^0=1\a_5=5^{5-4}=5^1=5[/latex] 2. a) [latex]a_n=frac{5n+2}{5n+3}=frac{5n+3-1}{5n+3}=frac{5n+3}{5n+3}-frac{1}{5n+3}=1-frac{1}{5n+3}\a_{n+1}=1-frac{1}{5(n+1)+3}=1-frac{1}{5n+8}\a_{n+1}-a_n=1-frac{1}{5n+8}-1+frac{1}{5n+3}=frac{-1(5n+3)+1(5n+8)}{(5n+8)(5n+3)}=\=frac{-5n-3+5n+8}{(5n+8)(5n+3)}=frac{5}{(5n+8)(5n+3)}>0[/latex] Ciąg jest rosnący b) [latex]b_n=frac{1}{3^n}\b_{n+1}=frac{1}{3^{n+1}}=frac{1}{3cdot3^n}\b_{n+1}-b_n=frac{1}{3cdot3^n}-frac{1}{3^n}=frac{1-3}{3cdot3^n}=frac{-2}{3cdot3^n}<0[/latex] Ciąg jest malejący 3. [latex]a_n=frac{n}{4}\frac{n}{4}=-8\n=-32 otin N_+\frac{n}{4}=0\n=0 otin N_+\frac{n}{4}=8\n=32\a_{32}=8[/latex] [latex]b_n=n^2+n-12\n^2+n-12=-8\n^2+n-4=0\Delta=1+16=17\n=frac{-1-sqrt{17}}{2} otin N_+ vee n=frac{-1+sqrt{17}}{2} otin N_+\n^2+n-12=0\Delta=1+48=49\n=frac{-1-7}{2} otin N_+ vee n=frac{-1+7}{2}=3\b_3=0\n^2+n-12=8\n^2+n-20=0\Delta=1+80=81\n=frac{-1-9}{2} otin N_+ vee n=frac{-1+9}{2}=4\b_4=8[/latex] 4. [latex]a_4=3\a_7=9\a_7-a_4=3r\3r=9-3=6\r=2\a_1=a_4-3r\a_1=3-3cdot2=3-6=-3\a_{30}=a_1+29r\a_{30}=-3+29cdot2=-3+58=55[/latex] [latex]S_{30}=frac{a_1+a_{30}}{2}cdot30\S_{30}=frac{-3+55}{2}cdot30=52cdot15=780[/latex]
