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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): l2 x: F$ y8 y
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Proof: 2 u. m" b6 Q+ e0 [' y
Let n >1 be an integer ; F5 S( f( q# c: F/ A9 `- p
Basis: (n=2)
! a" w% S9 a6 U/ Y8 y4 W) _! _" K 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3; x+ G# e/ C; j3 J, ]* w' i
/ q, ~; g7 f- M) }Induction Hypothesis: Let K >=2 be integers, support that
* ^1 `" V! y1 ]- B4 o# j K^3 – K can by divided by 3.5 Y" z/ j- |( i( C; d
& B5 A) v3 D, X, K* d: v4 c" wNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3$ _2 z, ?6 A2 {) ]8 H
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( K2 `" I& [, {
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)" {( C @5 y5 j3 b) o. H
= K^3 + 3K^2 + 2K
0 d4 n- ~. i. k4 D' S) z = ( K^3 – K) + ( 3K^2 + 3K)
0 n! x w( m s. _ l( h- n = ( K^3 – K) + 3 ( K^2 + K)
( g7 {. g5 F9 `2 ~: v% K! mby 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
; ?: B" V) t) u p! zSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# T* A2 V/ b1 |
= 3X + 3 ( K^2 + K); ~0 Q4 w( w9 k- C& D+ t) d% j2 f
= 3(X+ K^2 + K) which can be divided by 3 \1 f8 d3 Y; T0 s# D
; V$ L+ @. L+ u4 T4 T( EConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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! K. W7 D9 j8 y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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