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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)9 a" F' K0 T# i3 r
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Proof: / d1 b$ n6 Q( i+ r, L+ f5 o; _
Let n >1 be an integer
: p8 F+ P+ P. s* DBasis: (n=2)
1 O8 }/ C! E$ T: l 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that( J0 z$ X5 X& A( b0 r5 a
K^3 – K can by divided by 3.
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% {- S3 v e# b0 Y6 G) Y. YNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 Q& C+ K+ v- W2 m/ _1 Bsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: {! K @, h/ ? a) d; R' }# PThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 @% N8 c* x# r
= K^3 + 3K^2 + 2K3 w; B4 A# I9 @
= ( K^3 – K) + ( 3K^2 + 3K)+ ^) T1 x4 }- x4 D
= ( K^3 – K) + 3 ( K^2 + K)
% H. q2 ~( k, F- Fby 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
% j6 z) y9 D* XSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 D1 U2 q' z3 y. f* q n
= 3X + 3 ( K^2 + K)
8 Z/ G; g+ t6 c" J- ^2 \; r" G = 3(X+ K^2 + K) which can be divided by 3
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4 ~ _* o0 O3 U, \& eConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.$ A5 Q0 z6 B7 Q% y0 d, i+ f+ ?2 s
2 k) q* I4 K- e+ m4 ~" d" B# _; {/ T c[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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