数列{an}的前n项和Sn，Sn+an＝-1/2n²-3/2n+1(n∈N+） （1）设bn＝ a

n+n，证明：数列{bn}为等比数列
(2)若Cn=bn/(1-bn),数列cn的前n项和为Tn，证明Tn<5/3

Sn+an=-(1/2)n^2-(3/2)n+1
n=1
a1=-1/2
2Sn-S(n-1) = -(1/2)n^2-(3/2)n+1
2(Sn + (1/2)n^2 +(1/2)n -1) = S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1]=1/2
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S1 +(1/2)+(1/2) -1]=(1/2)^(n-1)
Sn + (1/2)n^2 +(1/2)n -1 = -(1/2)^n
Sn=1-n/2 -n^2/2 - (1/2)^n
an = Sn -S(n-1)
= -n +(1/2)^n
an +n = (1/2)^n
bn =an+n 是等比数列
nbn = n(1/2)^n
Tn =1b1+2b2+...+nbn
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=1/2
summation(i:1->n) i.(1/2)^(i-1)
= 4(n.(1/2)^(n+1) - (n+1).(1/2)^n + 1)
= 4[1- (n+2).(1/2)^(n+1)]

Tn =1b1+2b2+...+nbn
= (1/2)(summation(i:1->n) i.(1/2)^(i-1))
=2[1- (n+2).(1/2)^(n+1)]

an=4n²/(4n²-1）=1+1/(2n-1)(2n+1)

sn=n+[1/1×3+1/3×5+1/5×7……+1/（2n-1）（2n+1）]

=n+0.5[1-1/3+1/3-1/5+1/5-1/7……+1/（2n-1）-1/(2n+1)]

=n+0.5[1-1/(2n+1)]=n-1/(4n+2)+1/2

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