PROVE BY BINOMAIL THEOREM,
(8-1)^(2n)+8*2n-1 can be disivible by 64
BINOMAIL THEOREM
2007-09-28 3:46 am
回答 (2)
2007-09-28 4:09 am
✔ 最佳答案
(8-1)^(2n)+8*2n-1= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2* (-1)^(2n-2) + (8*2n)* (-1)^(2n-1)
+ (-1)^2n -1+8*2n-1
[Consider (8-1)^2n as (a-b)^2n, and then expand]
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2+ (-8*2n) + 1 - 1+8*2n-1
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
[because we take 64 as a factor out, so power of eight will be decreased by 2]
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
As those inside the bracket is integer,
(8-1)^(2n)+8*2n-1 can be disivible by 64
PS: you may simplify those in bracket if you like
2007-09-28 4:31 am
(8-1)^(2n)+8*2n-1
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2* (-1)^(2n-2) + (8*2n)* (-1)^(2n-1)
+ (-1)^2n -1+8*2n-1
[Consider (8-1)^2n as (a-b)^2n, and then expand]
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2+ (-8*2n) + 1 - 1+8*2n-1
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
[because we take 64 as a factor out, so power of eight will be decreased by 2]
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
As those inside the bracket is integer,
(8-1)^(2n)+8*2n-1 can be disivible by 64
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2* (-1)^(2n-2) + (8*2n)* (-1)^(2n-1)
+ (-1)^2n -1+8*2n-1
[Consider (8-1)^2n as (a-b)^2n, and then expand]
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2+ (-8*2n) + 1 - 1+8*2n-1
= 8^2n - (2n C 1) *8 ^(2n-1) + (2n C 2) * 8^(2n-2) + ... +
(2n C 2n-2) * 8 ^2
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
[because we take 64 as a factor out, so power of eight will be decreased by 2]
= 64 {[8^ (2n-2)] - (2n C 1) *8 ^(2n-3) + (2n C 2) * 8^(2n-4) + ... +
(2n C 2n-2) }
As those inside the bracket is integer,
(8-1)^(2n)+8*2n-1 can be disivible by 64
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