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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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- @9 N+ L7 c5 p( y* UProof:
" G. c# Z8 P# n$ eLet n >1 be an integer % M; h0 f. T! k5 h
Basis: (n=2)
4 @3 M4 r8 C( X9 Z4 c 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# |3 `' V2 y4 @9 t+ ^5 `
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Induction Hypothesis: Let K >=2 be integers, support that. @; k% Z7 Q9 R* @2 K( R
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 i; i( j/ Z2 Z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem/ A- T& A, m: u, \$ e0 H
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; Z0 t( b% I# B; T" Z = K^3 + 3K^2 + 2K5 K; i1 n$ X8 Q+ D( d R, J: o
= ( K^3 – K) + ( 3K^2 + 3K)
" L- {& H# e! G$ ? t% e = ( K^3 – K) + 3 ( K^2 + K)( n# _3 J% `' o2 S) l
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; k- I+ a Z- o& i& d1 Y" m1 @ E
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 e8 |* A! W: g* N = 3X + 3 ( K^2 + K)
& |2 S& P! X, h# N2 ~ = 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.# k _& N2 ~2 p
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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