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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: 9 O" R7 @$ m( L8 n7 K
Let n >1 be an integer
9 k1 r0 U7 E% Y! y7 cBasis: (n=2)
; w5 J5 w/ Z8 x+ g$ I) o4 ?# B 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: g6 w& z/ k) v1 {' p* |. y; J
" Z. G1 P) B+ N4 _: GInduction Hypothesis: Let K >=2 be integers, support that
" G0 e& _ [! ]7 q2 I% u K^3 – K can by divided by 3." g5 I6 T+ C* h; Y( o# N [
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
I: s. N: u" X% x; G Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
* I* P2 `0 j! p3 ]3 N zThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ @4 h* s/ X$ Y0 Y# Z- s = K^3 + 3K^2 + 2K
2 Y. T& G9 c& U/ `0 J2 T = ( K^3 – K) + ( 3K^2 + 3K)! L7 Y0 J3 _2 l0 Z( K4 X
= ( K^3 – K) + 3 ( K^2 + K)/ V3 o' ^$ a$ c* A" M
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
+ S0 r$ i# P, ?4 }# J1 r5 S" LSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 Y& L. H( t8 q1 p, Q- | = 3X + 3 ( K^2 + K)
+ \) c3 T! j" Z9 |0 T1 t% \ = 3(X+ K^2 + K) which can be divided by 33 F/ t6 L8 f8 m7 ] T$ v
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) h8 |% S0 a0 |: P
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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