Question

problema 6, va rog. e urgent :((((

Answer 1

Răspuns:

Folosim $\frac{1}{k\cdot (k+1)}=\frac{k+1-k}{k\cdot (k+1)}=\frac{1}{k}+\frac{1}{k+1}$

Atunci a = $\frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots +</p><p>\frac{1}{17}-\frac{1}{18} = \frac{1}{3} - \frac{1}{18} = \frac{5}{18}  = 0,2(7)$

Rezulta ca $a\in [0,(2) ; 0,(3)]$.

Answer 2

Răspuns:

1/3-1/4+1/4-1/5+...+1/16-1/17+1/17-1/18=1/3-1/18=6/18-1/18=5/18 ; restul s-au redus

Explicație pas cu pas:

o,(2)=2/9=4/18 ;2)  5/9=10/18;  4/18<5/18<10/18