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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) `: B) Z; L2 C: P c/ a1 b8 s) c
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Proof: 3 U" \# X% Y/ o: K- u1 g% |8 p* \$ F' ]
Let n >1 be an integer
1 Z8 P2 y8 G' P& OBasis: (n=2)1 b; S9 c+ J9 J1 u
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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/ f: K+ J- }1 ], V0 `Induction Hypothesis: Let K >=2 be integers, support that7 Q+ O/ E: A2 j& U% u
K^3 – K can by divided by 3.% K1 s$ V: m) B6 U
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 j& @. r) }$ k# s- q/ g8 _) Xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 A9 G* b3 |' p. W) ~3 cThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( E+ L1 M* m- j2 f6 w/ X" ~. q = K^3 + 3K^2 + 2K& I7 w2 |- x, O5 Q+ a
= ( K^3 – K) + ( 3K^2 + 3K)! Y3 r, A3 [7 k( h6 P( d
= ( K^3 – K) + 3 ( K^2 + K)
( v, M. J5 O# L p; q. ]) Q6 z9 x 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) [6 D$ w- D& L) ~
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 Q2 o4 | X- p( b" i = 3X + 3 ( K^2 + K)) q/ ^5 W7 `+ I! g% i( J
= 3(X+ K^2 + K) which can be divided by 3 p2 h5 q9 F4 I- _3 A9 v- r+ V
2 i( N) j/ Y" x1 L: a# ~Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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! j( E: F0 q# Z: N( _$ E# o4 Z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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