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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)' i0 q0 z3 x% @/ x/ F i
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Proof: 5 o0 D& ], v% v8 p
Let n >1 be an integer
2 w; D4 }/ I% O+ ] j# BBasis: (n=2)' z0 O( I6 C5 I: b$ m3 a5 l
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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2 U( q: E' d3 J8 Y u- o) HInduction Hypothesis: Let K >=2 be integers, support that
/ N. [0 p$ k/ S- E* H% D K^3 – K can by divided by 3.
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9 _1 n8 N1 G; V% \: D, {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% a2 ?1 r4 t" x0 K( i) ^5 Osince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 Q* [8 o2 ?2 [$ a6 f3 c v. m4 f4 sThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
8 ]9 }9 F3 Q% S( Q2 {! S: ` = K^3 + 3K^2 + 2K+ |5 x$ z) K6 f# Y% J
= ( K^3 – K) + ( 3K^2 + 3K)
% f) v( }. |0 K = ( K^3 – K) + 3 ( K^2 + K). V+ g" I( g5 Q, n2 h8 V: 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
. R0 Z' G7 y5 g" @' sSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); B ]: z; D6 b5 f
= 3X + 3 ( K^2 + K)% F# ~0 |2 k: P9 h, U* o
= 3(X+ K^2 + K) which can be divided by 31 Z V5 D$ f' j( N8 R# S! |
6 b. g* B) |, z5 A' h3 T7 f) C: pConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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