chyfbb
幼苗
共回答了10个问题采纳率:100% 举报
(1)设an=a1+(n-1)d,a2=a1+d,a4=a1+3d,lga1,lga2,lga4成等差数列,2lga2=lga1+lga4,
(a2)²=a1*a4,(a1+d)²=a1(a1+3d),得:a1=d,a2^n=d+(2^n-1)d=d2^n,则bn=1/a2^n=1/d2^n,{bn}首项为1/2d,公比q=1/2的等比数列;
(2)1/2d+1/4d+1/8d=7/24,d=3,数列{an}的首项a1=3,公差d=3;
(3)an=3+3n-3=3n,则{6anbn}=6n/2^n,Sn=6{1/2+2/2²+3/2³┄┄+n/2^n},上式两边乘以1/2得:1/2Sn=6{1/2²+2/2³+┄┄+n/2^(n+1)},Sn-1/2Sn=6{1/2+1/2²+2/2³+┄┄+1/2^n-n/2^(n+1)},
Sn=12{[1-1/2^n]-n/2^(n+1)}=12-(12n+6)/2^n
1年前
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