Răspuns:
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Explicație pas cu pas:
[tex]\dfrac{1}{4*5}+\dfrac{1}{5*6}+\dfrac{1}{6*7}+...+\dfrac{1}{23*24}=\\=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{23}-\dfrac{1}{24}=\dfrac{1}{4}-\dfrac{1}{24}=\dfrac{6}{24}-\dfrac{1}{24}=\dfrac{5}{24}\\\dfrac{1}{8}=\dfrac{3}{24},~~~\dfrac{1}{4}=\dfrac{6}{24}[/tex]
Intervalul (1/8; 1/4)=(3/24; 6/24). Deoarece 3/24<5/24<6/24, ⇒
5/24 ∈ (3/24; 6/24), deci 5/24 ∈ (1/8; 1/4).
