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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)2 X! V. o* L9 Q1 ~: u; A0 D9 A$ f& {
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Proof: : p. `4 w- i- L
Let n >1 be an integer
7 P/ ~; ]' B! Q# g% W7 ?& g% rBasis: (n=2)
7 q. F) E4 M. b7 B7 \* _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 Y2 P( P% H4 D2 l7 v# ^! Q
( @, L9 s" J" j0 aInduction Hypothesis: Let K >=2 be integers, support that$ _+ ?/ H0 H8 G$ _# @# F' }
K^3 – K can by divided by 3.
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' j5 R- C# w8 Z; h( e+ y1 H8 w- @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
) h" S' o5 s* R- }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 Z x; y0 s0 }$ c" B/ yThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 d2 b: [. Q) _5 @" Z
= K^3 + 3K^2 + 2K! j$ W% }) j- |1 X n
= ( K^3 – K) + ( 3K^2 + 3K)$ e/ m* @+ A5 f% }$ H* I
= ( K^3 – K) + 3 ( K^2 + K)5 }' x, n& z. p P5 K$ g9 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
; O( e- C% E( C* f( W' K3 V/ o; ZSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 u C9 {5 \) Z/ a% C = 3X + 3 ( K^2 + K): [( H$ h g, \4 g1 C6 y: G
= 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 >1.9 b' U* K: `$ |+ c! Z: O" N- ^
* F6 Q" ]# d. |9 H* ~# R. k# ^[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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