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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 b" i" G3 A1 L, [
% F6 X. \6 D+ D$ HProof:
6 [& ?1 p8 w9 o: VLet n >1 be an integer
8 _: B6 P$ A( t1 `) [Basis: (n=2)
0 F/ h9 Z- U& Q! w# ?# r/ U9 a* i* q6 E* l 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that C ?3 c3 x2 W: o* c3 e
K^3 – K can by divided by 3.
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5 Z7 I' K9 M/ [% E: _" R8 Z8 _Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ r# V8 {' {; Esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 ]5 R V `7 vThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)1 Y8 w7 V+ P+ Q: A2 f. `2 k- C
= K^3 + 3K^2 + 2K1 R8 n/ V/ _) c% p# ^7 Y, j
= ( K^3 – K) + ( 3K^2 + 3K), w9 L. v q! T
= ( K^3 – K) + 3 ( K^2 + K)
7 ~1 s5 B. u$ O' ?* iby 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>04 L# @5 I9 c4 n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
4 Y, Y9 U/ w6 W- w: W% ? = 3X + 3 ( K^2 + K)
9 m4 [& \/ {" d# J& j8 M- h4 j9 a = 3(X+ K^2 + K) which can be divided by 3- F5 {" u; d0 K: g* P4 o. u
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 A1 n- X+ V' G7 r$ _1 q4 @
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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