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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)0 L. L9 |3 s* l0 [0 H
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Proof:
# k1 m* _* R5 v) f9 pLet n >1 be an integer
$ U3 ], S7 s4 p6 T3 t5 s, @Basis: (n=2)
# C. Y1 I) B8 D* e8 Y9 U" ~ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 i9 _0 X& z5 x2 A' l1 {
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Induction Hypothesis: Let K >=2 be integers, support that; y" Y s q3 A; l
K^3 – K can by divided by 3.
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0 m$ p) c& s$ s4 cNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 d. ?2 f0 i, A0 X/ o, W4 C
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 Z. x* _3 h9 H+ ^5 |6 BThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 b& R" U" x, A- m5 a# P4 x
= K^3 + 3K^2 + 2K5 R8 e7 m6 }& D0 A/ X
= ( K^3 – K) + ( 3K^2 + 3K)! t3 ]0 y' ~) J6 w
= ( K^3 – K) + 3 ( K^2 + K)
, j" {% Q$ f& z% u; Pby 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 m/ H/ I% h* a9 E& I
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 s, s, r1 B5 _ s: u- M = 3X + 3 ( K^2 + K) e! h S' I ~6 c) I% G
= 3(X+ K^2 + K) which can be divided by 3% j" [' J4 E: _) k4 J C% j
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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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