作业帮已知数列an的前n项和为Sn,若Sn=2an+n,且bn=An－1/AnAn+1,求证an－1为等比数列;求数列{bn}
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已知数列an的前n项和为Sn,若Sn=2an+n,且bn=An－1/AnAn+1,求证an－1为等比数列;求数列{bn}的前n项和Tn
1、
n=1时,a1=S1=2a1+1
a1=-1
n≥2时,
Sn=2an+n S(n-1)=2a(n-1)+(n-1)
Sn-S(n-1)=an=2an+n-2a(n-1)-(n-1)
an=2a(n-1)-1
an-1=2a(n-1)-2=2[a(n-1)-1]
(an -1)/[a(n-1)-1]=2,为定值.
a1-1=-1-1=-2
数列{an -1}是以-2为首项,2为公比的等比数列.
2、
an -1=-2×2^(n-1)=-2ⁿ
an=1-2ⁿ
bn=(an -1)/[ana(n+1)]
=(1-2ⁿ-1)/[(1-2ⁿ)(1-2^(n+1))]
=1/(1-2ⁿ) -1/[1-2^(n+1)]
Tn=b1+b2+...+bn
=1/(1-2)-1/(1-2²)+1/(1-2²)-1/(1-2³)+...+1/(1-2ⁿ)-1/[1-2^(n+1)]
=-1 -1/[1-2^(n+1)]
n=1时,a1=S1=2a1+1
a1=-1
n≥2时,
Sn=2an+n S(n-1)=2a(n-1)+(n-1)
Sn-S(n-1)=an=2an+n-2a(n-1)-(n-1)
an=2a(n-1)-1
an-1=2a(n-1)-2=2[a(n-1)-1]
(an -1)/[a(n-1)-1]=2,为定值.
a1-1=-1-1=-2
数列{an -1}是以-2为首项,2为公比的等比数列.
2、
an -1=-2×2^(n-1)=-2ⁿ
an=1-2ⁿ
bn=(an -1)/[ana(n+1)]
=(1-2ⁿ-1)/[(1-2ⁿ)(1-2^(n+1))]
=1/(1-2ⁿ) -1/[1-2^(n+1)]
Tn=b1+b2+...+bn
=1/(1-2)-1/(1-2²)+1/(1-2²)-1/(1-2³)+...+1/(1-2ⁿ)-1/[1-2^(n+1)]
=-1 -1/[1-2^(n+1)]
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