设数列{an}的前n项和为Sn=4-1/4n-1(n属于N*),数列{bn}为等差数列,且b1=a1,a2(b2-b1)
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设数列{an}的前n项和为Sn=4-1/4n-1(n属于N*),数列{bn}为等差数列,且b1=a1,a2(b2-b1)=a1
(1)求数列{an}和{bn}的通项公式
(2)设cn=anbn,求数列{cn}的前n项和Tn
(1)求数列{an}和{bn}的通项公式
(2)设cn=anbn,求数列{cn}的前n项和Tn
Sn=4-1/4n-1,s(n+1)=4-1/[4(n+1)-1].后面减前面a(n+1)=1/(4n-1)-1/(4n+3)
于是an=1/(4n-5)-1/(4n-1)
a1=11/3.又a2(b2-b1)=a1.b2-b1=a1/a2=11/3/(27/7)=77/81.
bn=b1+(n-1)d=11/3+(n-1)*(77/81)=(77n+220)/81
于是an=1/(4n-5)-1/(4n-1)
a1=11/3.又a2(b2-b1)=a1.b2-b1=a1/a2=11/3/(27/7)=77/81.
bn=b1+(n-1)d=11/3+(n-1)*(77/81)=(77n+220)/81
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