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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)& b* V3 h2 y$ c& n# z( }- z1 M
# D) T9 \ j8 YProof:
, L& I* X' a: N" W: GLet n >1 be an integer
# F; o$ R: P2 _. ^; n' Y$ q+ m m1 fBasis: (n=2)' n% j5 ?$ V" w) s
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" z2 U$ N, }( n8 @+ o: ]* ]3 y
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Induction Hypothesis: Let K >=2 be integers, support that
1 E5 T/ [. X2 l. X3 L% @! R K^3 – K can by divided by 3.
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6 y' G: R" I+ gNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 i$ G2 {! G4 h. usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
4 _5 u/ W, g9 g$ t1 pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 c, f' D8 I: f2 {! c& g, d
= K^3 + 3K^2 + 2K
0 d: }: }8 S. x% u& {; v/ u2 R" \ = ( K^3 – K) + ( 3K^2 + 3K)
. J/ E$ p' f4 W = ( K^3 – K) + 3 ( K^2 + K)
q8 o3 l7 k5 S6 n: T, sby 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>04 N! n* f$ M; R+ T, P- ^
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 J2 b& l5 Z8 A# c l$ w3 n = 3X + 3 ( K^2 + K)
/ @0 b# G# I: h! \ = 3(X+ K^2 + K) which can be divided by 3
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% A9 \. n5 L/ X5 yConclusion: 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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