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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)0 V- Y* G0 A1 |/ J/ C4 l
; |' b5 k5 ^) \Proof: , w$ E! s6 U6 ] }4 ?
Let n >1 be an integer
3 S6 I! y0 x) a1 q* ` g" ]Basis: (n=2)4 h) C+ [9 r; d$ j$ m" Z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
* T8 k2 s! p M' l/ F' x' x5 s5 C+ ?- r8 J p& J
Induction Hypothesis: Let K >=2 be integers, support that
3 m$ b3 `4 f. I$ P4 Y# W 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 30 f9 t& B# Q" G
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, F6 g' F) w: m4 j0 t; I
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' p* x5 N" w5 B. c+ H, ^9 t" W# _" k = K^3 + 3K^2 + 2K% _0 N' a+ l6 T1 u: R, N0 h
= ( K^3 – K) + ( 3K^2 + 3K)
2 _& o7 y" b5 t8 A* ? = ( K^3 – K) + 3 ( K^2 + K)
( |* [2 a6 f. l/ U* gby 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
% B' z$ L; P7 T$ j. E- u! E0 ISo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)5 x1 K* B3 p# v, [
= 3X + 3 ( K^2 + K)
2 b2 O+ F2 L7 g) D% ^ = 3(X+ K^2 + K) which can be divided by 30 ^1 U5 p/ \8 {/ m
5 _& A5 P4 X2 }2 y
Conclusion: 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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