已知数列{an}满足a1+2a2+2^2a3+…+2^n-1an=n^2(n∈N*)(1)求{an}的通项公式(2)求{an}的前n项和Sn

1.
a1+2a2+2^2a3+…+2^n-1an=n^2 (1)
a1+2a2+2^2a3+...+2^(n-1)an+2^na(n+1)=(n+1)^2 (2)
(2)-(1)
2^n a(n+1)=2n+1
a(n+1)=(2n+1)/2^n
an=[2(n-1)+1]/2^(n-1)=(2n-1)/2^(n-1)
数列{an}的通项公式为an=(2n-1)/2^(n-1)
2.
an=(2n-1)/2^(n-1)=n/2^(n-2) -1/2^(n-1)
Sn=a1+a2+...+an=[1/2^(-1)+2/2^0+...+n/2^(n-2)] -[1/2^0+1/2^1+...+1/2^(n-1)]
令Cn=1/2^(-1)+2/2^0+...+n/2^(n-2)]
则Cn/2=1/2^0+2/2^1+...+(n-1)/2^(n-2)+n/2^(n-1)
Cn-Cn/2=Cn/2=1/2^(-1)+1/2^0+...+1/2^(n-2) -n/2^(n-1)
Cn=1/2^(-2)+1/2^(-1)+...+1/2^(n-3)-n/2^(n-2)
Sn=Cn -[1/2^0+1/2^1+...+1/2^(n-1)]
=1/2^(-2)+1/2^(-1)+...+1/2^(n-3)-n/2^(n-2) -[1/2^0+1/2^1+...+1/2^(n-1)]
=1/2^(-2) +1/2^(-1) -1/2^(n-2) -1/2^(n-1) -n/2^(n-2)
=6 -(2n+3)/2^(n-1)