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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 X5 A# U0 ~# |6 E. p
& Y) {5 i) D* RProof:
- p( H4 @' F* G3 KLet n >1 be an integer
5 Z( z; Y/ a& h1 T4 {Basis: (n=2)
F( s/ m5 q% ?8 E7 K! G 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
; a" S2 z8 q* v' G; g9 R) y+ M% d* g) n' d1 l% M
Induction Hypothesis: Let K >=2 be integers, support that
* n3 W' ?; a+ h- J, O5 l* l K^3 – K can by divided by 3.5 K, d7 S" j7 Z1 \& Q
1 }+ h3 A' ~' g! _; u
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ [* _/ z8 I) ~) ?! `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ C# M' t) c. {0 i ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, ?( t- }) p6 W7 d = K^3 + 3K^2 + 2K
4 m0 B, C4 H; i: [9 \0 [4 ?' C: N- N = ( K^3 – K) + ( 3K^2 + 3K)
& J, y& X9 Z, \7 L6 k+ | = ( K^3 – K) + 3 ( K^2 + K)
# S8 _8 r/ ^' }! xby 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
0 S' c4 k' m# o* {8 tSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ F3 w' y. B6 |5 i
= 3X + 3 ( K^2 + K)
; |1 c3 p: V3 B6 G = 3(X+ K^2 + K) which can be divided by 3! u! Y! ^, C0 Q
% N; l1 {* \2 T6 F2 ~Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; G, o' R3 E+ \; \$ x9 A' ^
4 D2 i: h: v5 J& c' l P" G& S[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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