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Answer 1

Răspuns:

$[\frac{1}{\sqrt{2} +\sqrt{1} } +\frac{1}{\sqrt{3} +\sqrt{2} } +...+\frac{1}{\sqrt{n+1} +\sqrt{n} }] =2$

$[\frac{\sqrt{2} -\sqrt{1}}{(\sqrt{2} +\sqrt{1})(\sqrt{2} -\sqrt{1})} +\frac{\sqrt{3} -\sqrt{2}}{(\sqrt{3} +\sqrt{2})(\sqrt{3} -\sqrt{2})} +...+\frac{\sqrt{n+1} -\sqrt{n}}{(\sqrt{n+1} +\sqrt{n})(\sqrt{n+1} -\sqrt{n})} ] =2$

$[\frac{\sqrt{2} -\sqrt{1}}{2-1} +\frac{\sqrt{3} -\sqrt{2}}{3-2} +...+\frac{\sqrt{n+1} -\sqrt{n}}{n+1-n} ] =2$

$[\sqrt{2} -\sqrt{1} +\sqrt{3} -\sqrt{2} +\sqrt{4} -\sqrt{3} +...+\sqrt{n} -\sqrt{n-1} +\sqrt{n+1} -\sqrt{n} ]=2$

$[\sqrt{n+1} -1 ]=2$

⇔ $2\leq \sqrt{n+1}-1 <3$

$3\leq \sqrt{n+1} <4$

⇔ 9 ≤ n+1 < 16

⇔ 8 ≤ n < 15

n ∈{8, 9, 10, 11, 12, 13, 14}