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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: 7 T: i, g4 A# @; \: C
Let n >1 be an integer
* e3 j4 X7 j9 \- o! i' oBasis: (n=2)
+ f* Y3 V7 I2 k b" J 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 that7 I( Z1 B P* z
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: b* ^8 b' O: {. l2 I
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem# V2 r% E' |& @$ x8 w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 O& Y. j* u4 G H5 ^
= K^3 + 3K^2 + 2K; Z$ _( M5 L( S& a0 H
= ( K^3 – K) + ( 3K^2 + 3K), s$ e3 s3 t1 ~& N4 _4 t6 l0 `
= ( K^3 – K) + 3 ( K^2 + K)
. ~) r5 i! p8 w; U" j8 tby 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>00 ], b. t! c, {/ X5 ~9 ^
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; S. V5 X+ o. ?( `" \8 C = 3X + 3 ( K^2 + K)
% m; S+ o! K2 X P5 L: v = 3(X+ K^2 + K) which can be divided by 3
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# F% V6 r* Z% s9 @+ H$ T vConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 A2 |/ ~' G2 I, i# t3 u
9 V: c# f- R6 K7 e+ a3 H( [[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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