Kaiken juuri

Paljonko on

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\cdots \frac{1}{\sqrt{24}+\sqrt{25}}?$

Ratkaisu: Koska laventamalla saadaan

$\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}=\frac{\sqrt{2}-\sqrt{1}}{2-1}=\sqrt{2}-1,$

ja edelleen

$\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2},$

ja koska vastaava lavennus toimii kaikille summan tekijöille, niin

$\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\cdots +\sqrt{25}-\sqrt{24}=5-1=4.$

Opettaja H:n pulmakulma