Question

设数列{An}的前n项和为Sn,已知S(n+1)=4An+2(n∈N*),A1=1,Bn=A(n+1)-2An.

Answer 1

题目不完整啊!

(1)A1=1,S2=A1+A2=4A1+2,得A2=5

S(n+1)=4An+2

Sn=4A(n-1)+2

S(n+1)-Sn=4An-4A(n-1)=A(n+1)

所以[A(n+1)-2An]/[An-2A(n-1)]=2

所以数列{A(n+1)-2An}是以A2-2A1=3为首项,2为公比的等比数列.

即A(n+1)-2An=3*2^(n-1)

所以Bn=A(n+1)-2An=3*2^(n-1)

2)Dn=An/2^n

D(n-1)=A(n-1)/2^(n-1)

Dn-D(n-1)=An/2^n-A(n-1)/2^(n-1)

=[An-2A(n-1)]/2^n

=3*2^(n-2)/2^n

=3/4

所以Dn是以A1/2=1/2为首项,3/4为公差的等差数列