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Concursul Matefbc editia a 4-a problema 1

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sun Dec 06, 2009 10:09 pm    Post subject: Concursul Matefbc editia a 4-a problema 1 Reply with quote

Se considera multimea X=\{1, 3, 5, 7, ....., 2n+1\}. Consideram urmatorul sir de submultimi ale multimii X astfel:
A_1=\{1\}; A_2=\{3, 5\}; A_3=\{7, 9, 11\}; etc.
a) Scrieti multimile A_4 si A_5;
b) Calculati suma elementelor multimii A_{20}.
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Alin
Euclid


Joined: 27 Mar 2010
Posts: 16

PostPosted: Sun Mar 28, 2010 3:13 pm    Post subject: Reply with quote

Punctul a) : \normal\ A_{4}=\left{13,15,17,19\right} si \ A_{5}=\left\{21,23,25,27,29\right}
Punctul b) : Ne punem problema care este primul element din multimea \normal\ A_{20} . Stim ca un numar impar se poate scrie sub forma \normal\2k+1. Astfel : \normal\ A_1=\left{2*0+1\right};\ A_2=\left\{2*1+1, 2*2+1\right};\ A_3=\left{2*3+1,2*4+1,2*5+1\right}; \ A_4=\left{2*6+1,2*7+1,2*8+1,2*9+1\right} samd. Se poate observa ca primul numar din fiecare multime are ca \normal\ indice (prin indice intelegem urmatorul numar de dupa inmultirea cu 2 si inainte de adunarea cu 1) suma cardinalelor multimilor de dinaintea lui. Spre exemplu primul element din \normal\ A_4 are primul element 2*6+1 cu indicele 6 ceea ce inseamna 1+2+3. Asa aplicam si pentru \ A_{20} . Primul element al sau este \normal\2*\left(1+2+3+\dots+19\right)+1, iar ultimul este \normal\2*(1+2+3+\dots+19)+1+2*19. Calculand suma lor : \normal\ S=2*\frac{19*20}{2}*20+1+3+\dots+39=7600+20*20=7600+400=8000 . Acum nu stiu cat de pe intelesul celor de clasa a 5-a am fost. Se mai putea deduce pentru o minte putin mai stralucita o formula pentru primul element din fiecare multime si anume n^2-n+1, unde n este indicele multimii respective.

Exercitiu : Folosind eventual indicatiile de mai sus, calculati suma elementelor multimii \normal\ A_{500} .
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