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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:
9 s7 N4 K7 q( y; {8 {Let n >1 be an integer
1 U2 v- G! g+ s+ j1 oBasis: (n=2)4 j/ Q$ F4 f) z8 e& Z* X; b/ w
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) @( [) Q+ `+ Z8 g( T# R* Y
( e z7 @+ I' {3 K% V I. C7 r% xInduction Hypothesis: Let K >=2 be integers, support that$ m- _' N U7 p! o) M7 u
K^3 – K can by divided by 3.
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: M: \, W* q3 J' nNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
( r$ J/ l- \% o* \) Msince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( q7 E: c8 C4 F$ ]. Z( e0 cThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. |# ?- Y$ Y* k; F7 I. k = K^3 + 3K^2 + 2K
( w7 ^9 X ]; O- E) U = ( K^3 – K) + ( 3K^2 + 3K)
" j$ z, Q! U! I/ k = ( K^3 – K) + 3 ( K^2 + K)& q" f2 y! W# x" _# J% G
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 v- \/ g( i3 O- O$ h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)2 z" T2 r* y3 y
= 3X + 3 ( K^2 + K)$ Z+ N; C6 i. ~3 v% b. D, n; `
= 3(X+ K^2 + K) which can be divided by 3" f4 j$ l- L& B, l
; `2 _+ D+ R9 x, M# [3 k# cConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) n7 G- r/ T+ @# w( Q& V' E5 ~
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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