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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)
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Proof:
" k5 ^. v: x5 l% X0 pLet n >1 be an integer 3 Y3 }& @( Z* U! a5 T8 N
Basis: (n=2)# \7 R& @7 e& |$ P R
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 o! |4 _9 e; K9 g2 o V5 W* s/ Y
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Induction Hypothesis: Let K >=2 be integers, support that
2 J" n- f6 W" O4 ` K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 T7 S( |8 {0 x# J1 I5 e
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% X* @5 q6 C7 @+ _/ eThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 x/ F3 Z% [4 I7 X
= K^3 + 3K^2 + 2K8 ^8 L. E. s" i z: F
= ( K^3 – K) + ( 3K^2 + 3K)
) ^' {% ~4 l/ M$ y = ( K^3 – K) + 3 ( K^2 + K)
& m3 w' p/ E( W3 S& x& _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 x4 W: R4 h$ i1 h P1 W8 ESo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! n$ K2 _; R% }, V& s
= 3X + 3 ( K^2 + K)4 i" R# P$ K8 J2 N" V8 L% j
= 3(X+ K^2 + K) which can be divided by 3* r! u8 X0 c$ ] ~/ H, l
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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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d% m# U! i! [4 ~[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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