Question

Arătați că pentru oricare "n" număr natural avem: [√n²+1]+ [√n²+2]+…+[√n²+2n]=2n².

Answer 1

$\displaystyle Observam~ca~n^2\ \textless \ n^2+1\ \textless \ n^2+2\ \textless \ ...\ \textless \ n^2+2n\ \textless \ (n+1)^2. \\ \\ Rezulta~n\ \textless \ \sqrt{n^2+1}\ \textless \ \sqrt{n^2+2}\ \textless \ ...\ \textless \ \sqrt{n^2+2n}\ \textless \ n+1. \\ \\ Deci~toate~expresiile~\sqrt{n^2+1},~\sqrt{n^2+2},...,~\sqrt{n^2+2n}~sunt~ \\ \\ situate~in~intervalul~(n,n+1). \\ \\ Rezulta~ca~fiecare~expresie~are~partea~intreaga~n. \\ \\ ~Deci~suma~data~este~\underbrace{n+n+...+n}_{de~2n~ori}=2n^2.$