Question

evaluate

$} \frac{1}{1*\sqrt{2}+2(\sqrt{1}) } +} \frac{1}{(2\sqrt{3} +3\sqrt{2})} +......\frac{1}{399\sqrt{400}+400\sqrt{399} }$

Answer 1

$\frac{1}{a\sqrt{b} +b\sqrt{a} } =\frac{a\sqrt{b} -b\sqrt{a} }{a^2b-b^2a} =\frac{a\sqrt{b} -b\sqrt{a} }{ab(a-b)} =\frac{\sqrt{b} }{b(a-b)} -\frac{\sqrt{a} }{a(a-b)}$

In the question, a-b=-1

$\frac{\sqrt{b} }{b(a-b)} -\frac{\sqrt{a} }{a(a-b)}=\frac{\sqrt{a} }{a} -\frac{\sqrt{b} }{b}$

If we put it in first terms

$\frac{\sqrt{a} }{a} -\frac{\sqrt{b} }{b} +\frac{\sqrt{b} }{b} -...$

We can observe : in the middle they all cancel out

The answer is $\frac{\sqrt{1} }{1} +\frac{\sqrt{400} }{400}$ or 1.05.

For your information these questions are called

telescope because the middle terms disappear

like a telescope.