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

1
se citeste 1 supra rad 2 + 1
daca reusesc sa o fac iau 10
la raspunsuri scrie √n-1=40 de unde rezulta ca n=1681
poate te ajuta
gata

Răspunsuri

2014-10-20T18:54:48+03:00
 \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 =>

 \sqrt{n} =40+1=41


Ridicam la patrat expresia, ca sa eliminam radicalul:

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

=> n= 1681