数列的问题1

已知等差数列{an}{bn}的前n项和为Sn,Tn.,m是任意正整数，（1） Sm/Sn=m^2/n^2求am/an.（2） am/an.=m/n,求Sm/Sn

热心网友

1.设Sm=km2,Sn=kn2am/an=(Sm-Sm-1)/(Sn-Sn-1)=k(m2-(m-1)2)/k(n2-(n-1)2)=(2m-1)/(2n-1)2.Sm/Sn=[(a1+am)m/2]/[(a1+an)n/2]=(a1+am)m/[(a1+an)n=[(1+am/a1)m]/[(1+an/a1)n]=[(1+m)m]/[(1+n)n]

热心网友

已知等差数列{an}{bn}的前n项和为Sn,Tn.,m是任意正整数，（1） Sm/Sn=m^/n^求am/an.（2） am/an.=m/n,求Sm/Sn{bn},Tn????解:（1）∵S2m-1=(a1+a2m-1)(2m-1)/2=am(2m-1)同理可得S2n-1=an(2n-1)∴S2m-1/S2n-1=am(2m-1)/an(2n-1)∴(2m-1)^/(2n-1)^=am(2m-1)/an(2n-1)am/an=(2m-1)/(2n-1)（2）Sm/Sn=[(a1+am)m/2]/[(a1+an)n/2]=(a1+am)m/[(a1+an)n=[(1+am/a1)m]/[(1+an/a1)n]=[(1+m)m]/[(1+n)n]