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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)4 c7 |4 T$ {- I* f3 W
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Proof: 8 k, Y/ C% A/ M8 N' ?! L
Let n >1 be an integer 6 W" ^: D2 n+ u; p- ^
Basis: (n=2)
! o6 g+ v' v& K+ u" _: e 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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; @: f4 ]5 J5 c; xInduction Hypothesis: Let K >=2 be integers, support that) @1 f! n; j* S/ f, ]) R
K^3 – K can by divided by 3.3 Q C8 u3 G4 H$ q
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 a+ C+ u/ P0 a# Ksince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem- F1 C" k1 ~* F
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. U4 p! D0 g5 ] = K^3 + 3K^2 + 2K
% e# E& Z0 v5 W# N! H1 a = ( K^3 – K) + ( 3K^2 + 3K)1 ~9 P' R5 C) G' t% W
= ( K^3 – K) + 3 ( K^2 + K)
n& U$ G2 J) Fby 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
$ |/ V4 a( a* |' q3 I1 [5 ?So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) R, y: w- [1 a8 N# m
= 3X + 3 ( K^2 + K)
2 Q! l+ ^& k! w = 3(X+ K^2 + K) which can be divided by 3
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+ p/ [3 `" `. h8 EConclusion: 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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