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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)# ^. I6 F9 j# z: ^( S
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Proof: ' v0 i R/ z8 {7 h. P" _, Y: X
Let n >1 be an integer
( S9 I+ S' ~7 f, q4 |Basis: (n=2)
9 R" z+ y; l& D9 e 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
% r) C% p9 v: e; _ K^3 – K can by divided by 3.
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% n7 n4 v9 @( s' o/ _$ h/ @; r6 G# K$ ~Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 39 I. N# N. _+ O7 g8 I# ^3 `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
a: i, u" Z) _! m: {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)2 p0 \2 b# v0 Z! @" r* w; W+ X
= K^3 + 3K^2 + 2K
0 O0 |) e- h, |! V = ( K^3 – K) + ( 3K^2 + 3K)# F! f1 | S, y% i0 L
= ( K^3 – K) + 3 ( K^2 + K)
% {0 @; k. |7 a4 X3 t! Zby 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% I, |' u' v; y3 L: V1 y) B
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
7 C6 L [2 {+ T+ q) _ = 3X + 3 ( K^2 + K)2 f( M' j+ A! n! |3 q* o
= 3(X+ K^2 + K) which can be divided by 3& r2 M6 h. o$ N& x
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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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