数列{an}满足a1+2a2+2^2a3+```````+2^(n-1)an=n^2(n属于整数)(1)求{an}的通项公式(2)求{an}的前n项和Sn

(1). a1=1^2=1
a1+2a2+2^2a3+```````+2^(n-2)a(n-1)+2^(n-1)an=n^2,
a1+2a2+2^2a3+```````+2^(n-2)a(n-1)=(n-1)^2,相减得:
an=(n^2-(n-1)^2)/2^(n-1)=(2n-1)/2^(n-1),n>1
{an}的通项公式:
an=(2n-1)/2^(n-1)
(2)..?

a1+2a2+2^2a3+```````+2^(n-1)an=n^2
a1+2a2+2^2a3+```````+2^(n-1)an+2^n×an=n^2+2^n×an
两个式子相减、2^n-1an-2^n-2an-1=1/2,即2^nan-2^n-1an-1=1，所以2^nan=2a1+（n-1)=n
所以an=n/2^n
s1=1/2
sn=...套公式，

1、a1+2a2+2^2a3+...+2^(n-2)a(n-1)+2^(n-1)an=n^2 (1)a1+2a2+2^2a3+...+2^(n-2)a(n-1)=(n-1)^2 (2)(1)-(2)2^(n-1)an=n^2-(n-1)^2=(n+n-1)(n-n+1)=2n-1an=(2n-1)/2^(n-1)=(4n-2)/2^n{an}的通项...