数列an,a1=1,an+a(n+1)=(1/4)^n,n是正整数.Sn=a1+4a2+4^2a3...+4^(n-1)an,求5Sn-4^nan=()

Question

数列an,a1=1,an+a(n+1)=(1/4)^n,n是正整数.Sn=a1+4a2+4^2a3...+4^(n-1)an,求 5Sn-4^n an=( )

Answer 1

Sn=a1+4a2+4^2a3...+4^(n-1)an
4Sn=4a1+4^2a2+4^3a3...+4^nan
5Sn=[a1+4a2+4^2a3...+4^(n-1)an]+[4a1+4^2a2+4^3a3...+4^nan]
=a1+{4(a1+a2)+4^2(a2+a3)+...+4^(n-1)*[a(n-1)+an]}+4^nan
=(1+4^nan)+{4*(1/4)^1+(4^2)*(1/4)^2+...+[4^(n-1)]*[(1/4)^(n-1)]}
=1+4^nan+(n-1)*1=4^nan+n
于是5Sn-4^nan=n