已知数列{an}的前n项和为Sn,且满足Sn=2an-2n(n∈N*)
2个回答
(1)
S1=a1=2a1-2
a1=2
Sn=2an-2n
Sn-1=2a(n-1)-2(n-1)
an=Sn-Sn-1=2an-2n-2a(n-1)+2(n-1)=2an-2a(n-1)-2
an=2a(n-1)+2
an +2=2a(n-1) +4
(an +2)/[a(n-1)+2]=2,为定值.
a1+2=2+2=4
数列{an +2}是以4为首项,2为公比的等比数列.
bn=an +2
数列{bn}是以4为首项,2为公比的等比数列.
bn=4×2^(n-1)=2^(n+1)
数列{bn}的通项公式为bn=2^(n+1)
(2)
cn=log2(bn)=log2[2^(n+1)]=n+1
cn/bn=(n+1)/2^(n+1)
Tn=c1/b1+c2/b2+...+cn/bn
=2/2^2+3/2^3+...+(n+1)/2^(n+1)
Tn/2=2/2^3+3/2^4+...+n/2^(n+1)+(n+1)/2^(n+2)
Tn-Tn/2=Tn/2=2/2^2+1/2^3+1/2^4+...+1/2^(n+1)-(n+1)/2^(n+2)
Tn=2/2+1/2^2+1/2^3+...+1/2^n -(n+1)/2^(n+1)
=1/2+1/2+1/2^2+1/2^3+...+1/2^n-(n+1)/2^(n+1)
=1/2 +(1/2)[1-(1/2)^n]/(1-1/2)-(n+1)/2^(n+1)
=3/2 -1/2^n- (n+1)/2^(n+1)
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