nicku92gm Posted March 8, 2011 Report Share Posted March 8, 2011 Suma primilor n termeni ai unei progresii aritmetice este S_n=5n^2+6n . Sa se determine termenul a_1 si ratia r Quote Link to comment Share on other sites More sharing options...
BogdanNBV Posted March 8, 2011 Report Share Posted March 8, 2011 asa, si ce ai rezolvat daca ai postat asta ? ne arati cum arata un exercitiu sau ? Quote Link to comment Share on other sites More sharing options...
Patrunjel Posted March 8, 2011 Report Share Posted March 8, 2011 1) si noi ce pula mea ar trebui sa facem cu asta?2) tin minte ca problemele de genu asta se rezolvau aplicand o forumla, pe care n-o mai tin minte.Daca te-ai uita in care te-ai prinde. Quote Link to comment Share on other sites More sharing options...
AlStar Posted March 8, 2011 Report Share Posted March 8, 2011 S_n=5n^2 +6n------------S_(n+1)-S_n=a_(n+1) => 5(n+1)^2 +6(n+1)-5n^2 -6n = a_(n+1) => 5(2n+1)+6 = a_(n+1) => a_(n+1) = 10n+11a_(n+1) = a_1 + nr => a_1 = 10n+11-nrS_n = (2*a_1 +(n-1)r)*n/2 = 5n^2 +6n => 2a_1 +(n-1)r = (10n^2 +12n)/n => 2a_1 + (n-1)r =10n +1220n+22-2nr+(n-1)r=10n+12 => 10n+10=r(2n-n+1) => 10(n+1)=r(n+1) => r=10a_1=10*1+11-1*10=11a_1=11 & r=10Mi-e somn... 1 Quote Link to comment Share on other sites More sharing options...
nicku92gm Posted March 8, 2011 Author Report Share Posted March 8, 2011 mersi mult alstar Quote Link to comment Share on other sites More sharing options...