Sa se arate ca pentru orice \( n\in\mathbb{N} \) , avem
\( \[\sqrt{n}+\sqrt{n+1}\]=\[\sqrt{4n+2}\] \).
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Observam ca \( \sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2} \).
Asadar \( [\sqrt{4n+1}]\le [\sqrt{n}+\sqrt{n+1}]\le [\sqrt{4n+2}] \).
Cum insa \( \sqrt{4n+2} \) nu poate fi patrat perfect, oricare ar fi n numar natural, notand \( [\sqrt{4n+2}]=k \), avem \( k^2<4n+2<(k+1)^2 \), adica \( k^2\le 4n+1<(k+1)^2\Longleftrightarrow k\le \sqrt{4n+1}<k+1 \), de unde \( [\sqrt{4n+1}]=k=[\sqrt{4n+2}] \).
Deci \( [\sqrt{4n+1}]=[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+2}] \).
Asadar \( [\sqrt{4n+1}]\le [\sqrt{n}+\sqrt{n+1}]\le [\sqrt{4n+2}] \).
Cum insa \( \sqrt{4n+2} \) nu poate fi patrat perfect, oricare ar fi n numar natural, notand \( [\sqrt{4n+2}]=k \), avem \( k^2<4n+2<(k+1)^2 \), adica \( k^2\le 4n+1<(k+1)^2\Longleftrightarrow k\le \sqrt{4n+1}<k+1 \), de unde \( [\sqrt{4n+1}]=k=[\sqrt{4n+2}] \).
Deci \( [\sqrt{4n+1}]=[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+2}] \).