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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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. _) r' G5 Y& k, @6 Z: g+ J7 gProof:
4 @8 t d& a, c3 u, ELet n >1 be an integer
& J2 M, N8 D& k- d" J: W4 j- Y9 EBasis: (n=2)+ t( k+ o/ C. d) o
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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! s0 M7 E8 A+ f$ B$ x" S; qInduction Hypothesis: Let K >=2 be integers, support that2 }' H- F8 S, D; O& _
K^3 – K can by divided by 3.& E B+ a9 ~. q' Y8 P) s
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( E% q+ ^$ P! L- X0 p, \
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 \5 G9 T2 t$ u u; j4 c/ bThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 c0 P5 N. e v+ d" D7 l l& O- R4 V
= K^3 + 3K^2 + 2K$ Z. o9 K4 }3 Z+ s$ Y8 Z- K5 r7 H
= ( K^3 – K) + ( 3K^2 + 3K)
4 ]! Z2 n3 x7 ?% o1 u( h1 f = ( K^3 – K) + 3 ( K^2 + K)
* R$ ^2 d* g4 H% x" L3 z8 Dby 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# \" ~* t- z% b6 K
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)6 q! Z) u/ y) d' h$ I
= 3X + 3 ( K^2 + K)
2 C1 w# _ Z% k8 r$ _+ z+ n = 3(X+ K^2 + K) which can be divided by 38 \2 [6 ~/ |6 v
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) l1 |( r) K0 c( N. z
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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