Question

Ma ajutati va rog?
Ex. 37

Answer 1

Vom amplifica cu conjugata fiecarui numitor, folosindu-ne de urmatoarea formula:
(a + b)(a - b) = a² - b²

[tex]k\in N^*\\\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}=\frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1})^2-(\sqrt{k})^2}=\frac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\\ =\sqrt{k+1}-\sqrt{k}[/tex]

Vom aplica formula demonstrata pentru fiecare fractie a sumei:

[tex]S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+..+\frac{1}{\sqrt{999}+\sqrt{100}}\\\\ S=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+...+(\sqrt{1000}-\sqrt{999})\\\\ S=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+...+\sqrt{999}-\sqrt{999}+\sqrt{1000}[/tex]

Se observa ca se reduc toti termenii cu semne opuse, mai putin primul si ultimul:
$S=\sqrt{1000}-\sqrt{1}=10\sqrt{10}-1$