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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)+ [* |/ I2 R: G, N
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Proof: $ [: W$ J" |3 V, |, l# e' f8 L5 ^
Let n >1 be an integer " N v# c4 k( l, H3 n7 W
Basis: (n=2)
5 `6 P* Z. F7 S" @ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ _ Y+ j1 K' p' l
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Induction Hypothesis: Let K >=2 be integers, support that
9 E: k; {- G; a1 Y K^3 – K can by divided by 3.# I& k' S% {4 l7 q$ b! Q! H
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 ?% Y, }( H) p0 Lsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 p4 p0 c; Y, ]' F, O6 O7 k6 NThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)/ h, b1 b3 N6 g) j1 y
= K^3 + 3K^2 + 2K+ T- ]/ C& U3 [/ ]: c! y
= ( K^3 – K) + ( 3K^2 + 3K)% \) W6 F* g+ ?9 u% f
= ( K^3 – K) + 3 ( K^2 + K), `/ e `0 o& D! U9 ?2 f0 h- C
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 D% f" b- T9 s* ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). Q7 D0 W4 P8 A9 {' w
= 3X + 3 ( K^2 + K)
) O/ ~( | R9 g/ L = 3(X+ K^2 + K) which can be divided by 39 U% W0 K3 E% G# m' D1 K, \1 ]
4 W. F' T$ d7 T% ]Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ ?2 L) Q" y$ X4 n5 {( E; P& p0 i
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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