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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)2 `0 x& b# a' k' R% j. Y
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Proof: ) D5 ^: `! e+ l: Z* z
Let n >1 be an integer
0 a' R; s5 h: x# IBasis: (n=2): D0 C7 R% o+ w4 i3 D
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' I. b- N* t0 W4 b
1 T# K/ K8 Z2 m/ C }* O
Induction Hypothesis: Let K >=2 be integers, support that4 X5 A+ @) F' E+ O8 E
K^3 – K can by divided by 3.9 n% g- d' d7 n
6 Z) ?2 ^5 x( vNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: l1 g% u; }* c V: k" z/ Y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ Q5 w) h( o4 n, N" Q) CThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
7 F" W: x/ Z; H$ W: c/ N; m = K^3 + 3K^2 + 2K) Y/ f! H0 T% D$ l0 h
= ( K^3 – K) + ( 3K^2 + 3K)! ^% q" `% }) l
= ( K^3 – K) + 3 ( K^2 + K)
) z( n2 q8 C" C; d |2 J1 {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
2 b7 n& u& u1 y+ S( ^, k% ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; q b. g( C5 |1 S( B' m! p1 z = 3X + 3 ( K^2 + K)
2 L- X& e) y" t9 E, h = 3(X+ K^2 + K) which can be divided by 3; o; D! M9 M& Q1 f9 f
+ S1 @! t, O# N3 ]+ `Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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