Question

已知数列{an},sn是它的前n项和,且s(n+1)=4an+2,a1=1.

(1)bn=a(n+1)-2an,求证:{bn}为等比数列.
(2)设cn=an/2^n.求证:数列{cn}是等差数列.

Answer 1

解:
(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)Cn=An/2^n
C(n-1)=A(n-1)/2^(n-1)
Cn-C(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D
Cn是以A1/2=1/2为首项,3/4为公差的等差数列