Question

Daca a si b sunt numere naturale nenule si

[tex]
\frac{ a^{n} }{ b^{n} } + \frac{ a^{n}+1 }{ b^{n}+1 } + \frac{ a^{n}+2 }{ b^{n}+2 } +....+ \frac{ a^{n}+2016 }{ b^{n} +2016} =2017,[/tex] demonstarti ca a=b, unde n aprtine N, fixat

Answer 1

I) daca a^n >b^n, atuci a^n =b^n +k , k e N,k>0=>
=> a^n/b^n +...+ (a^n+2016)/(b^n+2016)=
=(b^n+k)/b^n +...+(b^n +k+2016)/(b^n+2016)=
=b^n/b^n +k/b^n +...+(b^n+2016)/(b^n+2016)=
=1+1+1..+1 (de 2017 ori) + [k/b^n +...+k/(b^n+201 6) ] =
=2017+ [k/b^n +...+k/(b^n+201 6) ] >2017
II) daca a^n<b^n => a=b^n -k , k e N, k>0
............................................................................
=2017 -[k/b^n +...+k/(b^n+201 6) ]  <2017

din I) si II) => a^n=b^n, iar cum a,b,n e N => a=b