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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)8 b$ e" i2 ?+ L* Z9 C0 O) S
1 Q: u8 }9 f- i. D' J% F6 JProof: - Q( i, F5 S( G3 B
Let n >1 be an integer
0 Y$ W# Q9 M& A, Q2 f0 PBasis: (n=2)2 |7 L7 u% v4 ]) c$ o8 P1 a1 G
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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- i1 E9 n- t, L* P$ DInduction Hypothesis: Let K >=2 be integers, support that
; }3 ^8 E* T- Y4 j/ m K^3 – K can by divided by 3.
3 g. t' L0 F* _8 Q# p% s, C0 G2 o1 a# [, j1 ~; ^
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
# h1 n8 s7 w, m6 O( u- r( Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, n2 }( v; c9 P, iThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( `0 W& ?. d! }" K; M) V. s$ j = K^3 + 3K^2 + 2K
, u& J L- s) [0 J+ ]& R+ ` = ( K^3 – K) + ( 3K^2 + 3K)# M. [5 H3 d' k0 W `. ^3 q
= ( K^3 – K) + 3 ( K^2 + K)8 w& @6 G% ~* j
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& P0 p/ m: ]5 B: l
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( p& b5 j [. b# O = 3X + 3 ( K^2 + K)
2 H$ g4 I4 z! {/ R6 a = 3(X+ K^2 + K) which can be divided by 3
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4 L) ^0 \- [/ o: C, m& MConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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0 L, S+ b8 N8 m E9 J. u2 j[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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