已知{an(n为下标,同下)}是各项为不同正数的等差数列,lga1(1为a的下标,同下),lga2,lga4成等差数列,
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已知{an(n为下标,同下)}是各项为不同正数的等差数列,lga1(1为a的下标,同下),lga2,lga4成等差数列,又bn=1/a2^n(2^n为a的下标),n=1,2,3……
(1)证明:{bn}为等比数列
(2)如果数列{bn}的前三项和为7/24,求数列{an}的首项及公差
(3)在(2)小题的前提下,令Sn为数列{6anbn}的前n项和,求Sn
(1)证明:{bn}为等比数列
(2)如果数列{bn}的前三项和为7/24,求数列{an}的首项及公差
(3)在(2)小题的前提下,令Sn为数列{6anbn}的前n项和,求Sn
(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
(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
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