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OLM Prahova 2008
Posted: Thu Feb 19, 2009 12:56 am
by Marius Mainea
Se da sirul 1, 9,35,189 , 341 , 559 , 855,....
a) Enumerati urmatorii doi termeni ai sirului.
b) Aratati ca al 2008-lea termen al sirului este divizibil cu 5.
A. Apostol,Ploiesti
Posted: Thu Feb 19, 2009 2:11 pm
by Marcelina Popa
Daca regula gasita de mine e corecta, inseamna ca lipseste un termen, al patrulea:
1, 9, 35, 91, 189, 341, 559, 855, ...
In cazul asta urmatorii termeni ar fi 1241 si 1729.
E o problema "de fisa": iti cade fisa - iei punctajul maxim. Nu-ti cade, nu iei nici juma' de punct. Cam grea pentru faza locala, zic eu. Dar draguta!
Posted: Sun Feb 22, 2009 12:33 pm
by George+++
dar care e regula pana la urma?189=91*2+7
91=35*2+21
35=3*9+8..........................
Posted: Sun Feb 22, 2009 10:55 pm
by dede
\( a_1=1\,\, a_2=9 \,\, ... \,\, a_8=855 \)
\( b_i=a_{i+1}-a_i,\,\, i=1,\,\, 2,\,\, ... \)
\( c_i=b_{i+1}-b_i,\,\, i=1,\,\, 2,\,\, ... \)
\( Avem\,\, ca\,\, c_1=18,\,\, c_2=30,\,\, c_3=42 \,\, in\,\, general \,\, c_{i+1}=c_i+12 \,\,
atunci \,\, avem \,\,relatia\,\, a_{i+2}=a_{i+1}+b_i+c_i\,\, \\
deci\,\, a_9=855+296+90=1241\,\, si \,\,a_{10}=1241+386+102=1729 \\
Se \,\, observa \,\, ca \,\, 5/a_{5k+3} \,\, si \,\, 2008=5*401+3. \)
Posted: Mon Feb 23, 2009 12:49 am
by BogdanCNFB
\( a_1=1=0^3+1^3;a_2=9=1^3+2^3;a_3=35=2^3+3^3;...;a_n=(n-1)^3+n^3 \)
Deci, \( a_9=8^3+9^3=1241 \) si \( a_10=9^3+10^3=1729 \).
\( a_{2008}=2007^3+2008^3=(M_5+2)^3+(M_5+3)^3=M_5+2^3+M_5+3^3=M_5+35=M_5 \).