数列an的前n项和为Sn,Sn+an=-1/2n2-3/2n+1(n属于正自然数).设bn=an+n,证明数列bn是等比数列求数列｛nbn｝的前n项和Tn

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数列an的前n项和为Sn,Sn+an=-1/2n2-3/2n+1(n属于正自然数).设bn=an+n,证明数列bn是等比数列求数列｛nbn｝的前n项和Tn
数列an的前n项和为Sn,Sn+an=-1/2n2-3/2n+1(n属于正自然数).设bn=an+n,证明数列bn是等比数列
求数列｛nbn｝的前n项和Tn

数列an的前n项和为Sn,Sn+an=-1/2n2-3/2n+1(n属于正自然数).设bn=an+n,证明数列bn是等比数列求数列｛nbn｝的前n项和Tn
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)]

当n=1时，
S1+a1=2a1=-1/2-3/2+1=-1,
即a1=-1/2,
当n>=2时，
∵Sn+an=-1/2n^2-3/2*n+1,
∴S(n-1)+a(n-1)=-1/2(n-1)^2-3/2*(n-1)+1,
两式相减:
an+an-a(n-1)=-n-1/2
即2an-a(n-1)=-n-1,
∴...

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当n=1时，
S1+a1=2a1=-1/2-3/2+1=-1,
即a1=-1/2,
当n>=2时，
∵Sn+an=-1/2n^2-3/2*n+1,
∴S(n-1)+a(n-1)=-1/2(n-1)^2-3/2*(n-1)+1,
两式相减:
an+an-a(n-1)=-n-1/2
即2an-a(n-1)=-n-1,
∴2an=a(n-1)-n-1
∵bn=an+n=1/2a(n-1)+(1/2)n-1/2=1/2a(n-1)+1/2(n-1)
∴b(n-1)=a(n-1)+(n-1),
∴bn/b(n-1)=1/2
又b1=a1+1=1/2,
故bn是首项为1/2,公比为1/2的等比数列。
由(1)得：bn=1/2^n
∴｛nbn｝=n/2^n
∴Tn=1/2+2/2^2+3/2^3+……+n/2^n
∴2Tn=1+2/2+3/2^2+4/2^3+……+n/2^(n-1)
两式相减得：
Tn=1+1/2+1/2^2+1/2^3+……+1/2^(n-1)-n/2^n
=[1-(1/2)^n]/(1-1/2)-n/2^n
=2[1-(1/2)^n]-n/2^n
=2-(n+2)/2^n

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