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This can by done by Induction
Show that for all integers n >1, n^3 - n can be divided by 3. (Note: n^3 stands for n*n*n)4 x7 B2 ]! T7 R& a5 g
9 o: P: X7 C, m! V1 _( RProof: - J+ Y4 b- s7 P2 f0 P( v
Let n >1 be an integer % P. S/ L% S; E6 y
Basis: (n=2)
( O" ?! F0 C# ? 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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1 ~8 d) ?% E0 j6 H) }/ ~+ F7 ~Induction Hypothesis: Let K >=2 be integers, support that
[+ n& g5 w$ w+ ] K^3 – K can by divided by 3.8 _# {- {5 w4 x n( l
) V/ x w N) \: W
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" J1 Q* P, J: I6 I7 S# m" ^8 }: u- t
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! c: R* y' A' Q1 nThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, j* Z4 f1 g8 C8 g1 ]: y, w2 v2 a( { = K^3 + 3K^2 + 2K
) ]% J$ g7 Y5 L5 l i4 J = ( K^3 – K) + ( 3K^2 + 3K)8 A+ n7 |: k& K1 x4 `; {
= ( K^3 – K) + 3 ( K^2 + K)
* i5 s* ]) Q; u; l& R& ^6 S3 hby Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>03 c: _5 t7 H( ^% T3 k; p
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). W/ k6 P6 A0 {
= 3X + 3 ( K^2 + K)
8 } _; X% z/ |6 d = 3(X+ K^2 + K) which can be divided by 3$ b; P! |: }$ M! Z& U
! A8 w. Y. w3 h- PConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. @, Y: N4 z7 g! X& F8 l
3 B5 ^7 F( a, e* d; F/ v
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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