数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n
1个回答
1.
a(n+1)+1=(an²+2an+1)/(2an)=(an+1)²/(2an)
a(n+1)-1=(an²-2an+1)/(2an)=(an-1)²/(2an)
[a(n+1)+1]/[a(n+1)-1]=(an+1)²/(an-1)²
(an+1)²/(an-1)²>0,数列{(an+1)/(an-1)}各项均为正.
lg{[a(n+1)+1]/[a(n+1)-1]}=lg[(an+1)²/(an-1)²]=2lg[(an+1)/(an-1)]
bn=(an+1)/(an-1)
lg[b(n+1)]=2lg(bn)
lg[b(n+1)]/lg(bn)=2,为定值.
lgb1=lg[(a1+1)/(a1-1)]=lg[(2+1)/(2-1)]=lg3,数列[lg(bn)}是以lg3为首项,2为公比的等比数列.
2.
lg[(an+1)/(an-1)]=lg(bn)=lg3 ×2^(n-1)=lg{3^[2^(n-1)]}
(an+1)/(an-1)=3^[2^(n-1)]
an={3^[2^(n-1)]+1}/{3^[2^(n-1)]-1}
an-1=2/{3^[2^(n-1)]-1}
(an-1)/[a(n+1)-1]=[3^(2ⁿ)-1]/{3^[2^(n-1)]-1}
={3^[2^(n-1)]+1}{3^[2^(n-1)]-1}/{3^[2^(n-1)]-1} /分子用平方差公式
=3^[2^(n-1)]+1
3.
an=1+ 2/{3^[2^(n-1)]-1}
令cn=an-1
a2=(a1²+1)/(2a1)=(4+1)/4=5/4
n=2时,S2=a1+a2=2+ 5/4=13/4 2+4/3=10/3 13/4
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