Question

Determinati n∈N* astfel incat sa aiba loc egalitatea:
 1/√2+1   +   1/√3+√2   +   1/√4+√3   +....+1/√n+√n-1=40

Answer 1

$\frac{1}{ \sqrt{2}+1 } + \frac{1}{ \sqrt{3}+ \sqrt{2} }+\frac{1}{ \sqrt{4}+ \sqrt{3} }+...+\frac{1}{ \sqrt{n}+ \sqrt{n-1} } =40$



Analizam primele doua fractii:
$\frac{1}{ \sqrt{2}+1 } + \frac{1}{ \sqrt{3}+ \sqrt{2} }=$

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

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

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

$= \sqrt{3} -1$



Adaugam a treia fractie:
$\sqrt{3} -1 + \frac{1}{ \sqrt{4} + \sqrt{3} } =$

$= \frac{ (\sqrt{3} -1)( \sqrt{4} + \sqrt{3})}{\sqrt{4} + \sqrt{3}} + \frac{1}{ \sqrt{4} + \sqrt{3} } =$

$= \frac{ \sqrt{12}- \sqrt{4}- \sqrt{3} +4 }{ \sqrt{4} + \sqrt{3} } =$

$= \frac{( \sqrt{4}-1 )( \sqrt{4} + \sqrt{3} ) }{ \sqrt{4} + \sqrt{3} } =$

$= \sqrt{4} -1$



Adaugand fractie dupa fractie , pana la ultima gasim algoritmul:

$\sqrt{n} -1 =40$ =>

[tex] \sqrt{n} =40+1=41 [/tex]

Ridicam la patrat expresia, ca sa eliminam radicalul:

$( \sqrt{n} )^2=41^2$

=> $n= 1681$