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

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

解：
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-1
an=(2n-1)/2^(n-1)=(4n-2)/2^n
{an}的通项公式为an=(4n-2)/2^n
2、
Sn=a1+a2+...+an
=(4/2+4×2/2^2+4×3/2^3+...+4n/2^n)-2(1/2+1/4+...+1/2^n)
令Sn'=1/2+2/2^2+...+(n-1)/2^(n-1)+n/2^n
则Sn'/2=1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)
Sn'-Sn'/2=Sn'/2=1/2+1/2^2+1/2^3+...+1/2^n-n/2^(n+1)
=(1/2)[1-(1/2)^n]/(1-1/2)-n/2^(n+1)
=1-(1/2)^n-n/2^(n+1)
Sn'=2-1/2^(n-1)-n/2^n
Sn=4[2-1/2^(n-1)-n/2^n]-2(1/2)[1-(1/2)^n]/(1-1/2)
=8-2/2^n-4n/2^n-2+2/2^n
=6-4n/2^n

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，

n=1,a1=1/3 n>1, a1+3a2+3^2a3+……+3^(n-1)an=n/3① a1+3a2+3^2a3+……+3^(n-2)a(n-1)=(n-1)/3② ①-②得 3^(n-1)an=1/3, an=3^(-n), n=1时也符合 1)an通项为an=3^(-n) 2)bn=n*3^n sn=1*3^1+2*3^2+……+n*3^n③ 3sn=1*3^2+2*3^3+……+n*3^(n+1)④ ④-③得 2sn=n*3^(n+1)-(3^1+3^2+……+3^n)=n*3^(n+1)-3(3^n-1)/2=(n-1/2)*3^(n+1)+3/2 sn=(2n-1)*3^(n+1)/4+3/4