nicku92gm Posted March 8, 2011 Report 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
BogdanNBV Posted March 8, 2011 Report Posted March 8, 2011 asa, si ce ai rezolvat daca ai postat asta ? ne arati cum arata un exercitiu sau ? Quote
Patrunjel Posted March 8, 2011 Report 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
AlStar Posted March 8, 2011 Report 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