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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)/ m8 Q o k" g, |5 D) x
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Proof:
. k( B& [: i7 o. G6 ]! F* }9 i- wLet n >1 be an integer
- M5 H! }: ]2 _& JBasis: (n=2)
4 ]6 E9 w9 Y( g6 U+ P" _ u% I 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
2 G: t( i: M' l; `4 A! t" A0 |& v0 A$ X) O! }, x: L" ]0 F
Induction Hypothesis: Let K >=2 be integers, support that
7 q8 y; _5 B$ g3 r8 g1 c K^3 – K can by divided by 3.( r" C5 Q& \4 K$ I( e
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. ~" C# a, g9 E5 F( W) ^since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem$ U0 o# |( z8 M: ?4 m9 A [( ?7 b
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 R# Z! E7 V& S$ r# Z = K^3 + 3K^2 + 2K
% h( c9 T0 Q1 X9 x = ( K^3 – K) + ( 3K^2 + 3K)
3 ~& x, E- q- O) h = ( K^3 – K) + 3 ( K^2 + K)
5 }$ a1 q9 c( k0 Z" D* Tby 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% D L! r) C$ A/ n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
$ i3 D7 r, L6 x = 3X + 3 ( K^2 + K)! ] V u% \, X
= 3(X+ K^2 + K) which can be divided by 34 j* \3 \7 Q3 Q
: d! m0 }9 h) x$ L$ X
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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0 f$ U v% P. Q% M" S[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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