已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1/2成等差数列1求数列an的通项公式,若an方=2的负bn次方,设cn=bn/an,求数列cn的前n项和Tn
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![已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1/2成等差数列1求数列an的通项公式,若an方=2的负bn次方,设cn=bn/an,求数列cn的前n项和Tn](/uploads/image/z/10392850-10-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%2Can%2C1%2F2%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%971%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%2C%E8%8B%A5an%E6%96%B9%3D2%E7%9A%84%E8%B4%9Fbn%E6%AC%A1%E6%96%B9%2C%E8%AE%BEcn%3Dbn%2Fan%2C%E6%B1%82%E6%95%B0%E5%88%97cn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
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已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1/2成等差数列1求数列an的通项公式,若an方=2的负bn次方,设cn=bn/an,求数列cn的前n项和Tn
已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1/2成等差数列
1求数列an的通项公式,
若an方=2的负bn次方,设cn=bn/an,求数列cn的前n项和Tn
已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1/2成等差数列1求数列an的通项公式,若an方=2的负bn次方,设cn=bn/an,求数列cn的前n项和Tn
由题意
2an=Sn+1/2
Sn=2an-1/2
n=1时,S1=a1
a1=2a1-1/2
a1=1/2
S(n+1)-Sn=a(n+1)
2a(n+1)-1/2-[2an-1/2]=a(n+1)
a(n+1)=2an
因此{an}是等比数列,首项1/2,公比2
an=(1/2)*2^(n-1)
=2^(n-2)
Sn,an,1/2成等差数列
2an=1/2+Sn 2an-1=1/2+Sn-1
2a1=1/2+S1=1/2+a1 a1=1/2
2an-2an-1=1/2+Sn-1/2-Sn-1=an
an/an-1=2
所以{an}是以a1=1/2 q=2的等比数列
an=1/2×2^n=2^(n-1)
an^2=2^(-bn)
2^2(n-2)=2^(-bn)
bn=4-2n
cn=bn/an=(4-2n)/2^(n-2)
下面错位相减求和:
2cn=(4-2n)/2^(n-1)
……
2c1=(4-2*1)/2^0
Tn=(2c1+……+2cn)-(c1+……cn)
=2*[1/2^0+1/2^1+……+1/2^(n-2)]+(4-2n)/2^(n-1)-(4-2*1)/2^(-1)
=2*(1-1/2^(n-2))+(4-2n)/2^(n-1)-4
=(3-2n)/2^(n-1)-2