(1)
由于S(n+1)=4An+2
則有:Sn=4A(n-1)+2
兩式相減,得:
S(n+1)-Sn=4An-4A(n-1)
A(n+1)=4An-4A(n-1)
A(n+1)-2An=2[An-2A(n-1)]
由于Bn=A(n+1)-2An
則有:Bn=2B(n-1)
則:Bn/B(n-1)=2
則:數(shù)列{Bn}是公比為2的等比數(shù)列
(2)由于:
Bn
=B1*2^(n-1)
=(A2-2A1)*2^(n-1)
=(3/2)*2^n
則:
A(n+1)-2An=(3/2)*2^n
兩邊同時除以2^n得:
A(n+1)/2^n-2An/2^n=3/2
2[A(n+1)/2^(n+1)]-2[An/2^n]=3/2
由于Cn=An/2^n
則有:
2C(n+1)-2Cn=3/2
C(n+1)-Cn=3/4
則:數(shù)列{Cn}是公差為3/4的等差數(shù)列.