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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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9 Z7 r- w6 |) S1 g- C( s+ JProof: " h4 v# H+ \2 d- O3 G q& f
Let n >1 be an integer
( E6 M, J- s' r- t# W2 p6 DBasis: (n=2)5 `( L" u8 Z$ V' m
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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& `6 ]0 O, L0 b0 w# V7 @. KInduction Hypothesis: Let K >=2 be integers, support that
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 U9 @; c1 [9 @- b- ^; J4 |since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
) \2 G9 ~6 [7 Z3 xThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- v7 q: a1 H5 M& p3 W9 E = K^3 + 3K^2 + 2K s, t2 n: Y0 p8 d4 [) t
= ( K^3 – K) + ( 3K^2 + 3K)( D' e2 i& q8 a2 S# I1 ~# R
= ( K^3 – K) + 3 ( K^2 + K)
1 z7 N! a! x/ c! P4 _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
R- J% t) }- S+ q* {+ X. ]! t& XSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 V5 u1 n# b' b: I% o6 z% L$ O
= 3X + 3 ( K^2 + K)+ t% r2 F9 q8 I
= 3(X+ K^2 + K) which can be divided by 3' }/ b7 k) X" _/ M
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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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