[tex] \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[/tex]
Analizam primele doua fractii:
[tex] \frac{1}{ \sqrt{2}+1 } + \frac{1}{ \sqrt{3}+ \sqrt{2} }= [/tex]
[tex]= \frac{ \sqrt{3}+ \sqrt{2} }{( \sqrt{2}+1 )( \sqrt{3}+ \sqrt{2} )} + \frac{ \sqrt{2}+ 1 }{( \sqrt{2}+1 )( \sqrt{3}+ \sqrt{2} ) } =[/tex]
[tex] =\frac{ \sqrt{3} +2 \sqrt{2}+ 1 }{( \sqrt{2}+1 )( \sqrt{3}+ \sqrt{2} ) }=[/tex]
[tex]= \frac{ ( \sqrt{3} -1)( \sqrt{2}+1 )( \sqrt{3}+ \sqrt{2} ) }{( \sqrt{2}+1 )( \sqrt{3}+ \sqrt{2} ) } =[/tex]
[tex]= \sqrt{3} -1[/tex]
Adaugam a treia fractie:
[tex] \sqrt{3} -1 + \frac{1}{ \sqrt{4} + \sqrt{3} } =[/tex]
[tex]= \frac{ (\sqrt{3} -1)( \sqrt{4} + \sqrt{3})}{\sqrt{4} + \sqrt{3}} + \frac{1}{ \sqrt{4} + \sqrt{3} } =[/tex]
[tex]= \frac{ \sqrt{12}- \sqrt{4}- \sqrt{3} +4 }{ \sqrt{4} + \sqrt{3} } = [/tex]
[tex]= \frac{( \sqrt{4}-1 )( \sqrt{4} + \sqrt{3} ) }{ \sqrt{4} + \sqrt{3} } = [/tex]
[tex]= \sqrt{4} -1[/tex]
Adaugand fractie dupa fractie , pana la ultima gasim algoritmul:
[tex] \sqrt{n} -1 =40[/tex] =>
[tex] \sqrt{n} =40+1=41
[/tex]
Ridicam la patrat expresia, ca sa eliminam radicalul:
[tex]( \sqrt{n} )^2=41^2 [/tex]
=> [tex]n= 1681[/tex]
