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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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, l2 K! i; S% P+ X0 D! p( GProof: * K4 O0 K+ F n K5 s6 X
Let n >1 be an integer ' z- {9 M8 g( Y4 E: o0 F0 o3 K3 m' S
Basis: (n=2); g! L' [! X6 D2 X1 p3 R6 D) ]; S
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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6 W% j) `6 l* Y% gInduction Hypothesis: Let K >=2 be integers, support that
+ N# y+ N( P% N0 R0 F" X) D C K^3 – K can by divided by 3.! P/ @" W5 [/ _( |
8 K- C/ R# K" m/ r
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, `7 w, v/ ?& x+ q+ L1 ?since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
2 E1 ^7 J9 j: l8 W/ b( w( aThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)2 o: [/ H. A* B. C4 ~
= K^3 + 3K^2 + 2K7 E2 g, K; Z! e: _& H7 {, V- A/ @* @ M
= ( K^3 – K) + ( 3K^2 + 3K)8 u/ Y( A6 H( P X! \6 h
= ( K^3 – K) + 3 ( K^2 + K)
* }: p- _3 \) m; f, ~- S$ wby 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/ @1 T% g& t; O2 h; {/ f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)1 v4 E! M' s% u6 Z
= 3X + 3 ( K^2 + K)
- A. U: }, X' M# P0 J; f, f+ J = 3(X+ K^2 + K) which can be divided by 31 L( _; @. A1 C, B N1 j) z
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 H( Z4 R# A: U( T$ k
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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