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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:
. j/ @, z7 f$ K9 CLet n >1 be an integer + J" _( T% L/ k ^& c& h9 w& c
Basis: (n=2)* M9 Z% R" g/ r5 u; C1 U% ^" R
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. O6 G% `& _" X" h: [ c" F" t7 f
K^3 – K can by divided by 3.( b- w9 K4 y7 E! O/ j$ a
, i0 |/ G' K. p' O( j LNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ ~8 |, m! a" h
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, v% @" @% l& L/ \; F% W# `/ @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
# d$ y, a. ^. s# j7 Q; U- } = K^3 + 3K^2 + 2K2 C* ? t& `, j9 o5 ]+ m! F' H
= ( K^3 – K) + ( 3K^2 + 3K)
/ t Y2 w3 F( {5 r/ d = ( K^3 – K) + 3 ( K^2 + K)/ k. g' M8 g1 _) g6 }
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>08 j! K+ V3 x0 c" s& @/ n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)( [- M' @" N" s9 W$ G8 L
= 3X + 3 ( K^2 + K)8 p4 ]8 Y: {+ e2 G
= 3(X+ K^2 + K) which can be divided by 3# i$ [# U' n: N. E* `
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Conclusion: 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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