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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:
- Y5 w' O8 z+ C& _Let n >1 be an integer
4 x% g9 Y: T5 p+ O* Q3 q' N* MBasis: (n=2)
4 ^9 X9 K/ m& C0 c- @ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# S1 p9 ~1 M$ j1 ], B, t Y. V
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Induction Hypothesis: Let K >=2 be integers, support that3 \- e* m4 i. |) v
K^3 – K can by divided by 3.7 e, @5 b' D4 [% F
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 36 A8 i8 ]& L# j* q5 r5 }
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, e5 d$ x9 P; r$ P) ~6 K, V1 }Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1). V7 s* w6 i( u0 f
= K^3 + 3K^2 + 2K8 l: [ |4 L$ l- Q3 b7 S& y& O+ h" Z
= ( K^3 – K) + ( 3K^2 + 3K)& j& {5 P6 V& C! N j; c! i) O
= ( K^3 – K) + 3 ( K^2 + K)
) Z3 @. U0 W8 N, h9 [7 {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>06 B l# \) D( @" b5 U4 F$ j
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)7 V9 Q+ e4 E8 e+ l' A
= 3X + 3 ( K^2 + K)+ | i- P7 } h* z- q
= 3(X+ K^2 + K) which can be divided by 3& |) i- d+ c/ D w4 b2 c
4 s9 G. W) M8 T! A# h: _+ WConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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: _& q' Q5 O3 |; V, ]9 r[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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