已知数列an的前n项和为sn=n²+n 求通项公式,若bn=(2/1)an+n求数列bn的前

(1)
a1=S1=1^2+1=2
Sn=n^2+n
Sn-1=(n-1)^2+(n-1)
an=Sn-Sn-1=n^2+n-(n-1)^2-(n-1)=2n
{an}通项公式为an=2n
(2)
bn=(1/2)^an+n=(1/2)^(2n)+n=(1/4)^n+n
Tn=b1+b2+...+bn
=(1/4)^1+(1/4)^2+...+(1/4)^n+(1+2+...+n)
=(1/4)[(1-(1/4)^n]/(1-1/4)+n(n+1)/2
=1/3-(1/3)(1/4)^n+n(n+1)/2