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
问题描述：
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
答
(1)a1=S1=-a1-o.5^0+2=-a1+1a1=1/2a(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)^n2a(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+...