设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b(2n)-b(2n-1),求证cn前n项和Tn恒小于1.5
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![设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b(2n)-b(2n-1),求证cn前n项和Tn恒小于1.5](/uploads/image/z/1240662-30-2.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2Can%3D5sn%2B1+bn%3D%284%2Ban%29%2F%281-an%29%2C%E8%AE%B0cn%3Db%EF%BC%882n%EF%BC%89-b%EF%BC%882n-1%EF%BC%89%2C%E6%B1%82%E8%AF%81cn%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%E6%81%92%E5%B0%8F%E4%BA%8E1.5)
设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b(2n)-b(2n-1),求证cn前n项和Tn恒小于1.5
设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b(2n)-b(2n-1),求证cn前n项和Tn恒小于1.5
设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b(2n)-b(2n-1),求证cn前n项和Tn恒小于1.5
a1=-1/4
a(n+1)-an=[5S(n+1)+1]-(5Sn+1)
=5S(n+1)-5Sn
=5[S(n+1)-Sn]
=5a(n+1)
-4a(n+1)=an
a(n+1)=-1/4an
{an}是首项为-1/4,公比为-1/4的等比数列
an=(-1/4)^n
那么bn=(4+an)/(1-an)
=[4+(-1/4)^n]/[1-(-1/4)^n]
=[4-4(-1/4)^n+5(-1/4)^n]/[1-(-1/4)^n]=4+[5(-1/4)^n]/[1-(-1/4)^n]
后面那串上下同乘以(-4)^n 即得bn=4+5/[(-4)^n-1]
cn=b(2n)-b(2n-1)=4+5/[(-4)^2n-1]-{4+5/[(-4)^(2n-1)-1]}
=5/(16^n-1)-5/[-4^(2n-1)-1]=5/(16^n-1)+5/[4^(2n-1)+1](后面这串上下同乘以4)
=5/(16^n-1)+20/[4^(2n)+4]=5/(16^n-1)+20/[16^n+4](通分化简)
=25*16^n/[(16^n-1)*(16^n+4)]=16,所以(a-1)(a+4)>a²,分母越大,值越小}
T1=c1=25*16/(15*20)=4/3=2时,Tn