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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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6 v. d0 `9 z: `* p) fProof: 6 h1 I6 R, X7 W, P! |
Let n >1 be an integer
? V, w' ^0 Y h; |Basis: (n=2)
. i& q# H! b' D) H+ t 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ q% l# P1 _2 X2 K8 \' P! A
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Induction Hypothesis: Let K >=2 be integers, support that& k$ Z9 ]0 M% F! W" f' M: a) H
K^3 – K can by divided by 3.* p X0 w; y$ U
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. ?; u6 ]) v4 J' J5 D- esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 K0 V2 g$ g9 `! u4 ]. Q( D. e
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ U& ^) p8 b/ T7 \+ O; n6 o = K^3 + 3K^2 + 2K+ C. B9 g9 r) L. O- m. I( b
= ( K^3 – K) + ( 3K^2 + 3K)
/ H+ m6 x# @1 z2 X% J6 \) H8 O = ( K^3 – K) + 3 ( K^2 + K)
1 s4 T |4 P' Q5 \, g" M+ n4 B2 |8 b$ |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
1 o8 l' _: ^7 C9 q7 J Z' I6 G) J9 ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) ?) C" O! B& K' j$ T" d* b = 3X + 3 ( K^2 + K)
7 f/ c! S* W4 h( \5 s) {" t = 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.* i9 p+ Y9 `( X
) E0 e+ j8 W, m1 |( }- `6 n2 J[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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