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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)5 b6 V& y6 j: {1 I5 Y3 S
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Proof:
# Z) F. n8 y/ y+ k9 C: b8 w E1 _Let n >1 be an integer ; C2 k( {8 t" @/ X
Basis: (n=2), D6 v7 ]- q8 l
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, c' K3 K, P: s: a6 N, |1 r7 r, X: X* |
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Induction Hypothesis: Let K >=2 be integers, support that! I4 Z( N V2 T6 l# b
K^3 – K can by divided by 3.4 u( H4 D. k8 r# t; C
, E: v. t& D0 `7 ~) ^* HNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: D; \- `5 C% }( l
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: e) E: G: b- s
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); ~2 K( O4 l+ H: p0 J
= K^3 + 3K^2 + 2K
/ c ^( L. @/ u8 Z* y$ j. y; _ = ( K^3 – K) + ( 3K^2 + 3K)6 Z5 I0 \; Q4 H+ b1 f
= ( K^3 – K) + 3 ( K^2 + K)
: |7 [7 S+ L" y8 g5 nby 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( I+ c) J6 ~" x6 G; j8 O
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ d7 Z. W8 R$ C = 3X + 3 ( K^2 + K)
, t1 i5 C9 X4 f6 x = 3(X+ K^2 + K) which can be divided by 3% { N2 e3 V& R- N
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.9 i9 b5 W% v' B) _: o+ l# ]. N
. P3 N! J3 }4 q3 `5 u' L[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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