设数列{an}的前n项和Sn=3/2n2-1/2n,数列{bn}为等比数列,且a1=b1,b2(a2-a1)=b1,

问题描述:

设数列{an}的前n项和Sn=3/2n2-1/2n,数列{bn}为等比数列,且a1=b1,b2(a2-a1)=b1,
求数列{an},{bn}的通项公式

s1=3/2-1/2=1,s2=3/8-1/4=1/8 2a1+d=1/8s1=a1=1d=-15/8a2=-7/8 an=1-15(n-1)/8b1=a1=1b1q(-7/8-1)=b1q=-8/15 bn=(-8/15)n的次方