Question

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

Answer 1

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

Answer 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，

Answer 3

(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)..?