Răspuns:
ex.34;35
Explicație pas cu pas:
34. (la final, la numitor, este 100 în loc de 99)
[tex]x = \Big( \frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + 100} \Big) \cdot \frac{101}{100} = \\ [/tex]
[tex]= \Big( \frac{1}{1} + \frac{1}{ \frac{2 \cdot 3}{2} } + \frac{1}{ \frac{3 \cdot 4}{2} } + ... + \frac{1}{ \frac{100 \cdot 101}{2} } \Big) \cdot \frac{101}{100} \\ [/tex]
[tex]= \Big(1 + \frac{2}{2 \cdot 3} + \frac{2}{3 \cdot 4} + ... + \frac{2}{100 \cdot 101} \Big) \cdot \frac{101}{100} \\ [/tex]
[tex]= \Big[1 + 2 \cdot \Big(\frac{1}{2} - \not \frac{1}{3} + \not \frac{1}{3} - \not \frac{1}{4} + ... + \not \frac{1}{100} - \frac{1}{101} \Big)\Big] \cdot \frac{101}{100} \\ [/tex]
[tex]= \Big[1 + 2 \cdot \Big(\frac{1}{2} - \frac{1}{101} \Big)\Big] \cdot \frac{101}{100} = \Big(1 + 1 - \frac{2}{101} \Big) \cdot \frac{101}{100} \\ [/tex]
[tex]= \Big(2 - \frac{2}{101} \Big) \cdot \frac{101}{100} = \frac{200}{101} \cdot \frac{101}{100} \\ [/tex]
[tex]= \bf 2 \implies x \in \mathbb{N}\\ [/tex]
35.
știm că:
[tex] \bf \frac{1}{n} < \frac{n}{n + 1} < \frac{n + 1}{n + 2} \\ [/tex]
atunci:
[tex]\frac{1}{2} < \frac{2}{3} < \frac{3}{4} \\ [/tex]
[tex]\frac{3}{4} < \frac{4}{5} < \frac{5}{6} \\ [/tex]
[tex]\frac{5}{6} < \frac{6}{7} < \frac{7}{8} \\ [/tex]
....
[tex]\frac{49}{50} < \frac{50}{51} < \frac{51}{52} \\ [/tex]
înmulțim membru cu membru (pe verticală):
[tex]\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot ... \cdot \frac{49}{50} < \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot ... \cdot \frac{50}{51} < \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot ... \cdot \frac{51}{52} \\ [/tex]
q.e.d.
