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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:
: }8 S3 o8 a' a8 D% t( R2 ^5 D6 `Let n >1 be an integer 5 O9 L! X4 A* l( o/ }% \1 Q
Basis: (n=2)/ U- b+ Q" }- S& v2 \
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3- A# y [8 v$ [5 p; |' \' W
# F! _" f1 A& z: p. ] TInduction Hypothesis: Let K >=2 be integers, support that
0 }1 R9 |6 }% {' H" G+ S 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 35 @- t' {& ^+ |0 f. _" l+ S+ F
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% d* X* u& ~" L/ ^ @; Y# r7 HThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ L7 H$ d. Z" K V4 `: ~/ D6 L
= K^3 + 3K^2 + 2K* v8 d) V8 `* g
= ( K^3 – K) + ( 3K^2 + 3K)
) `- e, W& g/ |& F. F$ W3 g = ( K^3 – K) + 3 ( K^2 + K)
4 N3 d- B/ k, F. {( Jby 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
9 A/ U; y6 @9 w5 kSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)3 j/ O( _- r( ^ ]6 |& w
= 3X + 3 ( K^2 + K), {$ y( O6 z! e) J8 s% `+ _/ F
= 3(X+ K^2 + K) which can be divided by 35 A0 S. p# q" U% X+ D; [
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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& b/ \& T$ g, l[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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