已知等比数列{an}的前n项和为Sn,Sn=[(an+1)/2]²,n∈N*,bn=(-1)∧n×Sn,求数列bn的前n项和Tn.
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![已知等比数列{an}的前n项和为Sn,Sn=[(an+1)/2]²,n∈N*,bn=(-1)∧n×Sn,求数列bn的前n项和Tn.](/uploads/image/z/15208920-0-0.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%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%2CSn%EF%BC%9D%5B%EF%BC%88an%EF%BC%8B1%EF%BC%89%2F2%5D%26%23178%3B%2Cn%E2%88%88N%2A%2Cbn%EF%BC%9D%EF%BC%88%EF%BC%8D1%EF%BC%89%E2%88%A7n%C3%97Sn%2C%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn.)
已知等比数列{an}的前n项和为Sn,Sn=[(an+1)/2]²,n∈N*,bn=(-1)∧n×Sn,求数列bn的前n项和Tn.
已知等比数列{an}的前n项和为Sn,Sn=[(an+1)/2]²,n∈N*,bn=(-1)∧n×Sn,求数列bn的前n项和Tn.
已知等比数列{an}的前n项和为Sn,Sn=[(an+1)/2]²,n∈N*,bn=(-1)∧n×Sn,求数列bn的前n项和Tn.
n=1时,
a1=S1=[(a1+1)/2]²
(a1-1)²=0
a1=1
n≥2时,an=Sn-S(n-1)=[(an +1)/2]²-[(a(n-1)+1)/2]²
整理,得
an²-a(n-1)²-2an-2a(n-1)=0
[an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-2]=0
an=-a(n-1)或an-a(n-1)=2
an=-a(n-1)时,an/a(n-1)=-1,为定值,数列{an}是以1为首项,-1为公比的等比数列.
an-a(n-1)=2时,2为定值,数列{an}是以1为首项,2为公差的等差数列
an=1+2(n-1)=2n-1
a(n+1)/an=(2n+1)/(2n-1)=(2n-1+2)/(2n-1)=1+ 2/(2n-1)与n取值有关,不是等比数列,与已知不符.
综上,得数列{an}是以1为首项,-1为公比的等比数列.
Sn=1·[1-(-1)]ⁿ/[1-(-1)]=[1-(-1)ⁿ]/2
bn=(-1)ⁿ·Sn=(-1)ⁿ·[1-(-1)]ⁿ/2=[(-1)ⁿ-1]/2=(-1)ⁿ/2 -1/2
Tn=b1+b2+...+bn
=(-1)/2+(-1)²/2+...+(-1)ⁿ/2 -n/2
=(-1/2)[1-(-1)ⁿ]/[1-(-1)] -n/2
=[(-1)ⁿ-2n-1]/4