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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)/ m# l* T; T0 R0 z: D
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Proof: 8 w8 t2 a! ~( y4 m. C6 I; L
Let n >1 be an integer 9 m' Q: i( q' p
Basis: (n=2)+ H* \8 I. m& {* l
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
9 x2 j6 `7 W- v5 r# h' p! {3 H% h) G/ \4 v; O: ?$ ~; O5 [1 j& F" t
Induction Hypothesis: Let K >=2 be integers, support that
3 e- p" p% i9 G8 ~4 m5 V K^3 – K can by divided by 3.+ M; b4 {6 L. O1 [' j1 |
) @5 u+ }' }& @0 \: _; ^( iNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) F0 }$ j4 s. O- K( s' ~! P" l
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" ]& m- U/ L. _, b% I( ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 L% g) ]' S3 V1 i" z/ m
= K^3 + 3K^2 + 2K! w: j, H# j9 x3 C
= ( K^3 – K) + ( 3K^2 + 3K)
- R0 D3 e& H5 Z) ~ = ( K^3 – K) + 3 ( K^2 + K)
+ z1 E' s9 B5 v. a( M4 vby 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/ W2 {( R, n4 v6 q$ i
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 l w8 i2 O4 g = 3X + 3 ( K^2 + K)
1 ^7 a9 \7 `- f p, W& S, S = 3(X+ K^2 + K) which can be divided by 3
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& ?6 g N5 I$ @8 B; W$ A/ D6 a6 wConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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, v W- \: ~7 `4 X8 e& ^* s[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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