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
3个回答
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
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