Question

va rogggggggggggggg urgent​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Explicație pas cu pas:

$a = \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + \dfrac{1}{5 \cdot 6} + ... + \dfrac{1}{17 \cdot 18} = \dfrac{4 - 3}{3 \cdot 4} + \dfrac{5 - 4}{4 \cdot 5} + \dfrac{6 - 5}{5 \cdot 6} + ... + \dfrac{18 - 17}{17 \cdot 18} = \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{6} + ... + \dfrac{1}{17} - \dfrac{1}{18} = \dfrac{1}{3} - \dfrac{1}{18} = \dfrac{6 - 1}{18} = \dfrac{5}{18}$

$0,(2) = \dfrac{2}{9} = \dfrac{4}{18} < \dfrac{5}{18} < \dfrac{6}{18} = \dfrac{3}{9} = 0,(3)$

$\implies 0,(2) < a < 0,(3) \\ \iff a \in \Big[0,(2) ; 0,(3)\Big]$

q.e.d.