(1)
由于S(n+1)=4An+2
则有:Sn=4A(n-1)+2
两式相减,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
则有:Bn=2B(n-1)
则:Bn/B(n-1)=2
则:数列{Bn}是公比为2的等比数列
(2)由于:
Bn
=B1*2^(n-1)
=(A2-2A1)*2^(n-1)
=(3/2)*2^n
则:
A(n+1)-2An=(3/2)*2^n
两边同时除以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
则有:
2C(n+1)-2Cn=3/2
C(n+1)-Cn=3/4
则:数列{Cn}是公差为3/4的等差数列.