a)
[tex]{a_{n+1} - a_n =\frac{2(n+1)+3}{(n+1)+1} -\frac{2n+3}{n+1}=\frac{2n+2+3}{n+2} - \frac{2n+3}{n+1}=\frac{2n+5}{n+2} *\frac{(n+1)}{(n+1)} - \frac{2n+3}{n+1} *\frac{(n+2)}{(n+2)}=[/tex]
[tex]=\frac{(2n+5)(n+1)}{(n+2)(n+1)}-\frac{(2n+3)(n+2)}{(n+1)(n+2)}=\frac{2n^2+2n+5n+5-(2n^2+4n+3n+6)}{(n+1)(n+2)} =\frac{2n^2+7n+5-2n^2-7n-6}{(n+1)(n+2)}=\frac{-1}{(n+1)(n+2)} < 0[/tex]
poniewaz [tex](n+1)(n+2) > 0[/tex] dla kazdej n naturalnej
ciag malejacy
b)
[tex]a_{n+1}-a_n=\frac{3(n+1)+2}{(n+1)+4} -\frac{3n+2}{n+4} =\frac{3n+3+2}{n+5}- \frac{3n+2}{n+4} =\frac{3n+5}{n+5}*\frac{n+4}{n+4} - \frac{3n+2}{n+4} *\frac{n+5}{n+5}=[/tex]
[tex]=\frac{(3n+5)(n+4)-(3n+2)(n+5)}{(n+5)(n+4)}=\frac{3n^2+12n+5n+20-(3n^2+15n+2n+10)}{(n+5)(n+4)}=\frac{3n^2+17n+20-3n^2-17n-10}{(n+5)(n+4)}=\frac{10}{(n+5)(n+4)} > 0[/tex]
poniewaz (n+5)(n+4) >0 dla kazdej n naturalnej
ciag rosnacy
