Sa se demonstreze ca elementele multimii \( \{a,b,c\} \) sunt in progresie aritmetica, daca si numai daca este satisfacuta relatia:
\( a^2(b+c)+b^2(c+a)+c^2(a+b)=\frac{2}{9}(a+b+c)^2 \)
Progresii
-
Adriana Nistor
- Pitagora
- Posts: 82
- Joined: Thu Aug 07, 2008 10:07 pm
- Location: Drobeta Turnu Severin, Mehedinti
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Corect e asa : \( a^2(b+c)+b^2(c+a)+c^2(a+b)=\frac{2}{9}(a+b+c)^3\ \Longleftrightarrow\ \) elementele multimii \( \{a,b,c\} \) sunt in progresie aritmetica.
Relatia este echivalenta cu : \( 2a^3+2b^3+2c^3+12abc=3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \)
\( \Longleftrightarrow\ (a+b-2c)(b+c-2a)(c+a-2b)=0 \), deci concluzia rezulta cu usurinta.
Relatia este echivalenta cu : \( 2a^3+2b^3+2c^3+12abc=3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \)
\( \Longleftrightarrow\ (a+b-2c)(b+c-2a)(c+a-2b)=0 \), deci concluzia rezulta cu usurinta.