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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)
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Proof: 4 x* y0 p, ^0 [+ u6 h2 H9 W4 [
Let n >1 be an integer
9 L( e- N8 W( P cBasis: (n=2)5 u/ ~" j/ l# l i" {
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
) j: z4 O8 ?& P6 B; ^ K^3 – K can by divided by 3.
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) g9 _* H8 K( eNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, ?0 G0 g" \( M1 Ksince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 [$ d7 J5 Q0 @" _Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
& ?0 J8 Y8 [0 W8 i6 Q& M/ ^2 a = K^3 + 3K^2 + 2K
$ @& J1 `- v" K) W' T = ( K^3 – K) + ( 3K^2 + 3K)! }' {! h/ l6 v
= ( K^3 – K) + 3 ( K^2 + K)
5 ]) r' v* s! Cby 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
3 t4 o1 J( ]/ {0 ZSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)3 e h* i- m# M4 m9 c
= 3X + 3 ( K^2 + K)
8 a! ?4 U4 m) P' i6 G; A0 R$ X = 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ n* t I$ @7 h! ^7 J6 C0 G
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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