设数列{an}:a0=2,a1=16,a(n+2)=16a(n+1)-63an,n为正整数,则a2005被64除的余数为
设数列{an}:a0=2,a1=16,a(n+2)=16a(n+1)-63an,n为正整数,则a2005被64除的余数为16.
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a(n+2)=16a(n+1)-63an
The auxilary equation
x^2-16x+63=0
x=9 or 7
let
an = A(9)^n + B(7)^n
a0=2
A+B=2 (1)
a1=16
9A+7B=16 (2)
9(1)-(2)
2B=2
B=1
A=1
an = (9)^n + (7)^n
a2005 = 9^(2005) + 7^(2005)
9^(2005)
= (8+1)^2005
=2005C0(8)^2005 + 2005C1(8)^2004+ ...+ 1 (1)
7^(2005) =(8-1)^2005
=2005C0(8)^2005 - 2005C1(8)^2004+ ...- 1 (2)
(1)+(2)
9^(2005) + 7^(2005) = 2(2005C0.(8)^2005+2005C2.(8)^2003+...+8 )
[9^(2005) + 7^(2005)] mod 64
=[2(2005C0(8)^2005+2005C(8)^2003+...+8 )] mod 64
=2*8
=16
The auxilary equation
x^2-16x+63=0
x=9 or 7
let
an = A(9)^n + B(7)^n
a0=2
A+B=2 (1)
a1=16
9A+7B=16 (2)
9(1)-(2)
2B=2
B=1
A=1
an = (9)^n + (7)^n
a2005 = 9^(2005) + 7^(2005)
9^(2005)
= (8+1)^2005
=2005C0(8)^2005 + 2005C1(8)^2004+ ...+ 1 (1)
7^(2005) =(8-1)^2005
=2005C0(8)^2005 - 2005C1(8)^2004+ ...- 1 (2)
(1)+(2)
9^(2005) + 7^(2005) = 2(2005C0.(8)^2005+2005C2.(8)^2003+...+8 )
[9^(2005) + 7^(2005)] mod 64
=[2(2005C0(8)^2005+2005C(8)^2003+...+8 )] mod 64
=2*8
=16
1年前 追问
3
The auxiliary equation辅助方程
a(n+2)+ ca(n+1)+dan =0 let x1,x2 be roots of equation x^2+cx+d=0 ( auxiliary equation ) then an = A(x1)^n + B(x2)^n ( where A,B are constant )
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