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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), s! {& v O* W6 L) b% [( W
]5 Q. e) l- ?4 C ~8 m2 zProof:
$ J+ V# E% {. u' D3 u: wLet n >1 be an integer ) D4 L: z( j }) D6 k8 t+ C
Basis: (n=2)" M ]1 r9 m3 b# V7 h
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
; F' Q1 c1 y7 S1 g& g% W* ^, F1 a- f2 }: J
Induction Hypothesis: Let K >=2 be integers, support that1 ]. ~2 I) g( x" {/ j& S
K^3 – K can by divided by 3.
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+ X+ ?1 J% Y9 m7 M9 o9 MNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' ~" p D7 E6 a3 O
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: m; M/ i1 X; [* P9 O( [
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& @: a& _; y# ^; S; e
= K^3 + 3K^2 + 2K
( x; K, `- F5 T* V5 p = ( K^3 – K) + ( 3K^2 + 3K)
% I2 E) p+ k; z( D s! s = ( K^3 – K) + 3 ( K^2 + K)4 Y$ F* L6 t% N7 e& U
by 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>0
5 |, X1 Z A3 J$ |3 eSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)8 V" x- {6 p4 G! I
= 3X + 3 ( K^2 + K)
% |2 x8 j8 I6 P7 R: N) M = 3(X+ K^2 + K) which can be divided by 38 ]! }1 T5 o* @; i7 v3 ^9 j
4 G0 w. E7 q. E3 TConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.( k. o) x5 b3 Y j; H% v
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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