数列{an}满足a1+2a2+2^2a3+```````+2^(n-1)an=n^2(n属于整数)(

数列{an}满足a1+2a2+2^2a3+```````+2^(n-1)an=n^2(n属于整数)(

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}的通项公式为an=(4n-2)/2^n2、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^nSn=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======以下答案可供参考======供参考答案1:a1+2a2+2^2a3+```````+2^(n-1)an=n^2a1+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^ns1=1/2 sn=...套公式，供参考答案2:(1). a1=1^2=1a1+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)..?

感谢回答，我学习了