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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 X8 I6 J, S- ~$ k$ J
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Proof:
. O$ k3 G( \+ N5 \Let n >1 be an integer
! s5 L9 I1 X' ?$ n, i& x5 d( Z& q0 DBasis: (n=2)8 O4 w; B: T. |# s5 B
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
* E* G. S* v9 }/ P K^3 – K can by divided by 3.5 y/ }* k* r9 X, l* H6 X( m" w- _
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 C1 K& A: G2 V6 t4 O s+ Y4 xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 s& q8 r- S; a9 P% j2 q3 Q4 j: V7 S
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): [9 ?# o* i' @. K" \8 Q2 w
= K^3 + 3K^2 + 2K9 g: i8 i5 ^: P3 K/ \5 j
= ( K^3 – K) + ( 3K^2 + 3K)* e8 S. @; @ M8 i }3 t' b
= ( K^3 – K) + 3 ( K^2 + K)0 u& q# C/ I: K* r; L* M
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
: |3 s) i" T4 d3 } ISo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 h) c8 `) j- Y* u# C& w2 o = 3X + 3 ( K^2 + K)
+ r& l, L$ j$ ^8 d# Y5 {6 M = 3(X+ K^2 + K) which can be divided by 3
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' d$ t2 n6 D$ U9 t4 W5 J" bConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 U0 {( }" P! |) G+ v. v
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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