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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)4 E& x2 c4 m N/ \+ a" @3 k
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Proof:
n: Y. d* f6 h* J0 O h4 N nLet n >1 be an integer 9 u# ?+ ~( U, s/ u4 H! V
Basis: (n=2)
$ |. v- A$ |: A" m: r- u2 q h9 J" p 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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2 r( K0 x5 q { `Induction Hypothesis: Let K >=2 be integers, support that( W# N6 j! X& V9 q8 O9 j: @
K^3 – K can by divided by 3.
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1 [1 i% V a. i z: RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 e& @% M8 g6 l* B: I& ~+ h4 }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
* {) ]( }0 B ?. @$ @7 ~/ q4 sThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 v% ?3 w! G/ q( }5 [! k7 W = K^3 + 3K^2 + 2K
o* ?2 C' k% [( ^7 J = ( K^3 – K) + ( 3K^2 + 3K)# _4 d/ ? ^" V* p, r
= ( K^3 – K) + 3 ( K^2 + K)
* _5 o% l1 I2 M: B& aby 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 b- f# U" ~; z+ m3 p* h' L
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ w3 X ^0 @- L! t) j5 i
= 3X + 3 ( K^2 + K)# K" |: a$ V6 P& k. j6 }$ s5 d% }
= 3(X+ K^2 + K) which can be divided by 3. `/ w- K' }2 x& x) a6 S
) `% l& G G" k2 X9 q5 CConclusion: 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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