已知数列an的前n项和Sn=-an-(0.5)^n-1+2,(n为正整数),（1）令bn=2^nan,求证数列bn是等差

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已知数列an的前n项和Sn=-an-(0.5)^n-1+2,(n为正整数),（1）令bn=2^nan,求证数列bn是等差数列,并求数列an的通向公式（2）令cn=（n+1）an/n,Tn=c1+c2+···+cn,试求Tn

(1)a1=S1=-a1-o.5^0+2=-a1+1
a1=1/2
a(n+1)=S(n+1)-Sn
=[-a(n+1)-(1/2)^n+2]-[-an-(1/2)^(n-1)+2]
=-a(n+1)+an-(1/2)^n+(1/2)^(n-1)
=-a(n+1)+an+(1/2)^n
2a(n+1)=an+(1/2)^n
令bn=2^nan
则b(n+1)=2^(n+1)a(n+1)
b(n+2)=2^(n+2)a(n+2)
那么2b(n+1)=2×2^(n+1)a(n+1)
=2^(n+1)×2a(n+1)
=2^(n+1)[an+(1/2)^n]
=2^(n+1)an+2
bn+b(n+2)=2^nan+2^(n+2)a(n+2)
=2^nan+2^(n+1)×2a(n+2)
=2^nan+2^(n+1)[a(n+1)+(1/2)^(n+1)]
=2^nan+2^(n+1)a(n+1)+1
=2^nan+2^n×2a(n+1)+1
=2^nan+2^n[an+(1/2)^n]+1
=2^nan+2^nan+1+1
=2^(n+1)an+2
即：2b(n+1)=bn+b(n+2)
∴数列{bn}是等差数列
b1=2a1=1
b2=2^2a2=4a2=2[a1+(1/2)^1]=2
公差为2-1=1
通项为bn=1+(n-1)=n
那么2^nan=n
∴数列{an}的通项公式为：an=n/2^n
(2)令cn=(n+1)an/n
∵an=n/2^n
an/n=1/2^n
∴cn=(n+1)/2^n
Tn=c1+c2+c3+……+cn
=2/2^1+3/2^2+4/2^3+……+(n+1)/2^n
1/2Tn=2/2^2+3/2^3+4/2^4+……+n/2^n+(n+1)/2^(n+1)
Tn-1/2Tn=2/2^1+1/2^2+1/2^3+1/2^4+……+1/2^n-(n+1)/2^(n+1)
1/2Tn=1/2+(1/2^1+1/2^2+1/2^3+1/2^4+……+1/2^n)-(n+1)/2^(n+1)
1/2Tn=1/2+1-1/2^n-(n+1)/2^(n+1)
Tn=3-2/2^n-(n+1)/2^n
Tn=3-(n+3)/2^n

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