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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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& k, W/ X" N3 _3 q7 ]4 [# kProof: . o7 d5 \; C2 T$ T6 a! [! h2 ~
Let n >1 be an integer
* Z/ O: f5 W, a0 i7 l, cBasis: (n=2)4 Z/ h& ^ n7 h! h. t
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 that6 |1 O {+ }1 c# h
K^3 – K can by divided by 3.* o s" v7 f/ I) F$ ^; j
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 d. l" e* M! o
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ t, c' j1 d' }# v0 I2 S ]3 [" CThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, C( o# Y0 w9 J! ` = K^3 + 3K^2 + 2K
% I% c) j4 [- W1 t6 e" z$ } = ( K^3 – K) + ( 3K^2 + 3K)
' I" R2 B _) v+ Q) I = ( K^3 – K) + 3 ( K^2 + K)
3 d# |' e) O" ~3 v4 Q7 Cby 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
& V, L- t! q. }" U+ U" d CSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)8 y! Q$ y6 O, Q" X
= 3X + 3 ( K^2 + K)
" }& P! s* X1 ]0 d& L5 Y5 u+ A = 3(X+ K^2 + K) which can be divided by 3: ~0 Z- @0 p/ W
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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