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1. a) Dem. ca \( (n+3)^2-(n+2)^2-(n+1)2+n^2=4 \), oricare ar fi n\( \in \) N.
b) Aratati ca putem alega convenabil semnele \( "+" \) si\( "-" \) a.i., oricare ar fi n\( \in \) N sa aiba loc egalitatea:
\( \pm(n+7)^2\pm(n+6)^2\pm(n+5)^2\pm(n+4)^2\pm(n+3)^2\pm(n+2)^2\pm(n+1)^2\pm n^2=0 \)
c) Fie A={2010,2011,...,2025}. Aratati ca exista multimile \( B \) si \( C \) disjuncte (\( B\cap C=\emptyset \)) a.i. \( B\cup C=A \), suma elmentelor din B este egala cu suma elementelor din C si suma patratelor elementelor lui B este egala cu suma patratelor elementelor lui C.

a) \( (n+3)^2-(n+2)^2-(n+1)^2+n^2=n^2+6n+9-n^2-4n-4-n^2-2n-1+n^2=9-4-1=4 \).
b) Facem urmatoarea alegere a semnelor: \( (n+7)^2-(n+6)^2-(n+5)^2+(n+4)^2-(n+3)^2+(n+2)^2+(n+1)^2- n^2=[(n+4)+3]^2-[(n+4)+2]^2-[(n+4)+1]^2+(n+4)^2-[(n+3)^2-(n+2)^2-(n+1)^2+n^2]=4-4=0 \).
c)Observam ca multimea are 16 elemente. Din punctul b) deducem ca \( (n+7)^2+(n+4)^2+(n+3)^2+n^2=(n+6)^2+(n+5)^2+(n+2)^2+(n+1)^2 \).
De asemenea, \( (n+15)^2+(n+12)^2+(n+11)^2+ \)\( (n+\8\)^2 \)\( =(n+14)^2+(n+13)^2+(n+10)^2+(n+9)^2 \).
Alegem \( B=\{n, n+3, n+4, n+7, n+9, n+10, n+13, n+14\} \) si \( C=\{n+1, n+2, n+5, n+6, n+8, n+11, n+12, n+15\} \), unde \( n=2010 \).
Aceste doua multimi indeplinesc conditiile din ipoteza:
\( (n+7)^2+(n+4)^2+(n+3)^2+n^2+(n+14)^2+(n+13)^2+(n+10)^2+(n+9)^2=(n+6)^2+(n+5)^2+(n+2)^2+(n+1)^2+(n+15)^2+(n+12)^2+(n+11)^2+ \)\( (n+\8\)^2 \)
\( (n+7)+(n+4)+(n+3)+n+(n+14)+(n+13)+(n+10)+(n+9)=(n+6)+(n+5)+(n+2)+(n+1)+(n+15)+(n+12)+(n+11)+(n+\8\)=8n+60 \).