1.
[tex]a_{n} = \frac{2n+1}{n+3}\\\\a_{n+1} = \frac{2(n+1)+1}{(n+1)+3} = \frac{2n+2+1}{n+1+3} = \frac{2n+3}{n+4}\\\\a_{n+1}-a_{n} = \frac{2n+3}{n+4}-\frac{2n+1}{n+3} = \frac{(2n+3)(n+3)-(2n+1)(n+4)}{(n+4)(n+3)}=\frac{2n^{2}+6n+3n+9 -2n^{2}-8n-n-4}{(n+4)(n+3)}=\\\\=\frac{5}{(n+4)(n+3)} > 0 \ - \ ciag \ rosnacy[/tex]
Uzasadnienie:
[tex]n + 4 > 0 \ - \ dla \ n \in N+\\\\n+3 > 0 \ - \ dla \ dowolnego \ n \in N+\\\\(n+4)(n+3) > 0 \ - \ iloczyn \ liczb \ dodatnich \ jest \ liczba \ dodatnia\\\\\frac{5}{(n+4)(n+3)} > 0 \ - \ iloraz \ liczb \ dodatnich \ jest \ liczba \ dodatnia[/tex]
2.
[tex]a_{2} = 5\\a_{7} = 23\\a_{n} = ?\\\\a_{7} = a_2 + 5r\\\\23 = 5 + 5r\\\\23-5 = 5r\\\\5r = 18 \ \ /:5\\\\r = 3,6[/tex]
[tex]a_2 = a_1 + r\\\\5 = a_1 + 3,6\\\\5-3,6 = a_1\\\\a_1 = 1,4[/tex]
[tex]a_{n} = a_1 + (n-1)\cdot r\\\\a_{n} = 1,4+(n-1)\cdot3,6\\\\a_{n} = 1,4 + 3,6n - 3,6\\\\a_{n} = 3,6n-2,2[/tex]
