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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 q; d/ `: A) O8 o! k- o. `
Let n >1 be an integer ]" L6 m* n5 L1 V0 C7 K2 r! C( v
Basis: (n=2)
9 r4 [; W& N3 T) B 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 Z$ t' H+ s/ ?& h2 e! q, V
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Induction Hypothesis: Let K >=2 be integers, support that( I) s& E3 E6 H* h
K^3 – K can by divided by 3.5 U7 O8 D$ I) b
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 f( ~) Z m& R+ f, o# z" Ssince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, ~% j7 y+ c I6 c7 v
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
6 ^( o! @. z1 | = K^3 + 3K^2 + 2K/ V5 y! h2 C/ Q
= ( K^3 – K) + ( 3K^2 + 3K)6 p% G4 G/ q( ~
= ( K^3 – K) + 3 ( K^2 + K)5 G7 N8 W0 b# F/ `. H- U0 v
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
: n& g" s- ^6 GSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 |4 \6 r: w. ~- U. J2 ?) Z = 3X + 3 ( K^2 + K)
3 A5 ^2 e, Q& |: m: {5 t = 3(X+ K^2 + K) which can be divided by 3
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) C, R$ a- \7 d' v. KConclusion: 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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