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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)) j0 p5 ?" \& F& q ]. Q
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Proof:
- d2 Z- A3 q$ {Let n >1 be an integer % X P- J( t& L6 m' p3 z
Basis: (n=2)7 R+ H" r( w1 N1 Q
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! x5 X/ V0 ~0 X) h
K^3 – K can by divided by 3.: h9 X: D; s( J. O) h
2 J' K# z- d" o7 d# hNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ g; F6 W6 j# D4 d' }/ z! n- rsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, |0 u2 Z1 @- P y, `8 z" b
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)# k# I* e0 [1 G' t% c, t
= K^3 + 3K^2 + 2K- S3 ` a6 g6 K5 v' U; F8 B+ ^
= ( K^3 – K) + ( 3K^2 + 3K)
4 _) r1 m% }% t = ( K^3 – K) + 3 ( K^2 + K)
, b, L& H* q# ]; d/ E) h/ K1 |* B% V6 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
L# t2 k; j I( RSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" @0 _: F- z0 _; f2 l) B = 3X + 3 ( K^2 + K)! g \, Y4 X9 a1 G
= 3(X+ K^2 + K) which can be divided by 3
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* y( S, b5 k: q, [& F/ {Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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; v/ c2 o# J Y6 W, L[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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