Question

数列{An}的前n项和为Sn,且满足Sn=n-An,n属于正整数，（1）求证数列{An-1}是等比数列，

（2)设Cn=-2nAn+2n ，数列{Cn}的前n项和为Tn,求证：Tn<4

Answer 1

a(1)=s(1)=1-a(1), a(1)=1/2.
s(n+1)=(n+1)-a(n+1),
s(n) = n - a(n),
a(n+1)= s(n+1)-s(n)=[(n+1)-a(n+1)]-[n-a(n)]=1-a(n+1)+a(n),
2a(n+1) = 1 + a(n),
2a(n+1)-2=2[a(n+1)-1]=a(n)-1,
a(n+1)-1 = [a(n)-1]/2,
{a(n)-1}是首项为a(1)-1=-1/2,公比为1/2的等比数列。
a(n)-1 = -(1/2)^n,
a(n) = 1 - 1/2^n.

c(n) = 2n-2na(n)=2n[1-a(n)]=2n/2^n = n/2^(n-1),
t(n) = 1/1 + 2/2 + 3/2^2 + ... + (n-1)/2^(n-2) + n/2^(n-1),
2t(n) = 2/1 + 2/1 + 3/2 + ... + (n-1)/2^(n-3) + n/2^(n-2),
t(n) = 2t(n) - t(n) = 2 + 1 + 1/2 + ... + 1/2^(n-2) - n/2^(n-1)
=2 - n/2^(n-1) + [1-1/2^(n-1)]/[1-1/2]
=2 - n/2^(n-1) + 2[1-1/2^(n-1)]
=4 - (n+2)/2^(n-1)
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