已知两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且?
已知两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且 Sn/Tn = (7n+1)/(4n+27),则an/bn=?
设an=a1+x(n-1),bn=b1+y(n-1) 则Sn=n(a1+an)/2,Tn=n(b1+bn)/2; 展开得Sn/Tn=(2a1-x+xn)/2b1-y+yn) 由于Sn/Tn=(7n+1)/(4n+27) 知2a1-x+xn=7n+1,且x=7 ;2b1-y+yn=4n+27,且y=4 2a1-7=1得a1=4;2b1-4=27得b1=31/2; 所以an=7n-3,bn=4n+23/2 an/bn=2(7n-3)/(8n+23)