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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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! b3 g4 s) ^8 JProof:
5 q' m7 Y! E) Y4 {# n1 |2 HLet n >1 be an integer
9 G/ k n+ ~# [9 u$ Z' K6 b/ eBasis: (n=2)
1 Q" |2 z9 l% x. f+ {+ t9 g 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
1 t+ W! I/ I0 B& U" j9 Z$ }" q/ N( H; t& v, f+ |9 ?& I
Induction Hypothesis: Let K >=2 be integers, support that' l7 o# _' i p6 e! {3 c
K^3 – K can by divided by 3.
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6 \3 g2 v6 T$ YNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) L9 c- a+ O: i2 v7 V/ I4 z r
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
) `2 v; I7 V* Y2 HThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' C$ `4 k$ ?" U& e# r = K^3 + 3K^2 + 2K
- Y$ Y' `0 I* Q0 {0 Z0 z. U; { = ( K^3 – K) + ( 3K^2 + 3K)& Z2 @0 j5 c8 L2 E/ i4 z+ \/ o
= ( K^3 – K) + 3 ( K^2 + K)
: }4 z4 w$ n7 g: d9 q+ }5 ^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
4 X/ P: ]+ I8 @/ m) RSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)0 b1 f6 T5 k" k/ |
= 3X + 3 ( K^2 + K)
( {: s" j% B" I1 R = 3(X+ K^2 + K) which can be divided by 3
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$ Q$ G A- i- s7 rConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 j& U' `& b; w* T |* F5 X
$ D; V: s3 j- Q4 L+ q# _) @[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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