1、数列{an} 前n 项和Sn=-an-(1/2)的（n-1）次方+2（n为正整数）

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1、数列{an} 前n 项和Sn=-an-(1/2)的（n-1）次方+2（n为正整数）
令cn=(n+1/n)an,tn=c1+c2+c3+...+cn,试求Tn值
2、正项数列{an}前n项和Sn,方程x平方+4x-4Sn=0,有一根为an-1,求Tn=(1/S1)+(1/S2)+(1/S3)...+(1/Sn)
求证：Tn

答：
(1)
n=1时算出a1=1/2.
Sn=-an-(1/2)^(n-1)+2.
S(n+1)=-a(n+1)-(1/2)^n+2.
两式相减,
a(n+1)=an-a(n+1)+(1/2)^n.
2a(n+1)=an+(1/2)^n.
2^(n+1)×a(n+1)=2^n×an+1.
设Dn=2^n×an,则D1=2a1=1.
D(n+1)-Dn=1.
所以
Dn=n.
即2^n×an=n.
an=n/2^n.
cn=(n+1/n)an=(n+1)/2^n.
tn=2/2+3/2^2+...+(n+1)/2^n.
2tn=2+3/2+...+(n+1)/2^(n-1).
两式相减,
tn=2+1/2+1/2^2+1/2^(n-1)-(n+1)/2^n.
=3-(n+3)/2^n.
(2)
x^2+4x-4Sn=0.
an-1是方程的解.
(an-1)^2+4(an-1)-4Sn=0.
(an+1)^2=4(1+Sn).
n=1时算出a1=3.(an>0)
(an+1)^2=4(1+Sn).
[a(n+1)+1]^2=4[1+S(n+1)]
两式相减,
4a(n+1)=[a(n+1)+1]^2-(an+1)^2
=[a(n+1)]^2+2a(n+1)-(an)^2-2an.
[a(n+1)]^2-2a(n+1)-(an)^2-2an=0.
[a(n+2)+an][a(n+1)-an-2]=0.
{an}是正项数列.
所以
a(n+1)-an=2.
所以
an=2(n-1)+3=2n+1.
Sn=n^2+2n.
1/Sn
=1/(n^2+2n)
=1/2[1/n-1/(n+2)]
=1/2×(1/n)-1/2×[1/(n+2)]
Tn
=(1/S1)+(1/S2)+...+(1/Sn)
=1/2(1+1/2+...+1/n)-1/2[1/3+1/4+...+1/(n+2)]
=1/2+1/4-1/2[1/(n+1)+1/(n+2)]
=3/4-1/2[1/(n+1)+1/(n+2)]

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