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
人气:176 ℃ 时间:2019-09-09 11:37:53
解答
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
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