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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) G1 l: _- a; V. L l; ?/ A
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Proof: ( A6 @$ b* ?2 L7 |' P9 p8 c
Let n >1 be an integer 6 i$ a: [, ]4 e
Basis: (n=2)
2 @6 c' ]& A( U+ X% h' U3 w! E( T( c 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3- a8 m6 J. Q/ J+ f/ |, v+ X
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Induction Hypothesis: Let K >=2 be integers, support that+ { R0 G# P2 c2 {, C( ^
K^3 – K can by divided by 3.7 o5 I8 z1 @# y
: r6 }' B- s7 Q) T2 w. ]4 T. E* MNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
; c! A1 M. X5 f$ Z$ b8 u1 k, y }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; f! H6 J: Y9 F* T4 lThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
& v1 u5 ]5 z+ @+ G8 _1 r' i' p = K^3 + 3K^2 + 2K
H. l* O$ k l2 W5 a" ] = ( K^3 – K) + ( 3K^2 + 3K)
- s8 A z2 R4 P6 U U) y: q2 i = ( K^3 – K) + 3 ( K^2 + K)
) D+ D# n1 j& [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
* R9 P! c y- F" _* cSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). B. J' U- h, o! [0 @
= 3X + 3 ( K^2 + K)
" X! d) k3 {# B8 } = 3(X+ K^2 + K) which can be divided by 34 B, E+ Y9 v; e0 a; G2 `
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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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