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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: : v( m* b* v2 e! S5 \
Let n >1 be an integer " i- D' J7 G% l9 w- X
Basis: (n=2)3 f% k3 p; L4 F# K
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 that3 d- j- V1 H3 c' ^2 j
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 36 J T3 F0 v" m! `, r
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! @0 i8 ~# `0 `8 ~1 UThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' ^! B: X/ h' Y7 r% Q7 J& W: n
= K^3 + 3K^2 + 2K
' A. i5 | ?2 b U9 o = ( K^3 – K) + ( 3K^2 + 3K)* z9 O1 ]# v7 Y+ L ^0 n2 R5 c
= ( K^3 – K) + 3 ( K^2 + K)
F! G: [' Q+ l1 l) Lby 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
+ h) q9 m3 Y2 @0 H) F6 CSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* b* J; D# T/ G = 3X + 3 ( K^2 + K)
% J' O5 z% D2 U( T/ ^. B' L8 _ = 3(X+ K^2 + K) which can be divided by 3: @7 M4 l$ M# C6 p" [' d
2 R4 q7 b( w& I6 M: `! b0 }Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 |7 G+ e' o( d; d7 v1 _! Y
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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