设数列{an}:a0=2,a1=16,an+2=16an+1-63an,n∈N*,则a2005被64除的余数为( ) A.0 B.2 C.16 D.48
问题描述:
设数列{an}:a0=2,a1=16,an+2=16an+1-63an,n∈N*,则a2005被64除的余数为( )
A. 0
B. 2
C. 16
D. 48
答
∵an+2=16an+1-63an,
∴an+2-7an+1=9an+1-63an,或an+2-9an+1=7an+1-63an
即an+2-7an+1=9(an+1-7an),或an+2-9an+1=7(an+1-9an),
又∵a0=2,a1=16,
∴{an-7an-1}是首项为2,公比为9的等比数列,
{an-9an-1}是首项为-2,公比为7的等比数列.
∴an+1-7an=2•9n,
an+1-9an=-2•7n.
联立上述两式解得,
an=9n+7n,
a2005=92005+72005
=(8+1)2005+(8-1)2005
=2(
82005+
C
02005
82003+…+8),
C
22005
∴上式中除最后一项8之外都是64的倍数,
∴a2005被64除的余数为2×8=16.
故选:C.