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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)8 [9 o2 ~$ G7 X) ^# p6 r$ ]
# \5 ]2 b* Y( cProof:
& L' U; Y' z6 z B2 ?" JLet n >1 be an integer % Q" m1 H: [2 I1 X) C9 x
Basis: (n=2)
2 X+ J* {/ K2 l 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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/ Y( t4 ~8 B v) s* NInduction Hypothesis: Let K >=2 be integers, support that
5 l1 a2 q* J5 O, y+ @' r K^3 – K can by divided by 3.! y0 C8 W+ Q; T1 H4 N3 d. `
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( @6 m- s) r; g% _0 D2 v6 }, D. z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. `1 I" N5 _* d$ |Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% {* O& [" Z0 g6 Z" b- k = K^3 + 3K^2 + 2K
2 y: r- G8 r- C' j = ( K^3 – K) + ( 3K^2 + 3K)9 e) B9 e/ O7 J- {# m! B9 n& {: e
= ( K^3 – K) + 3 ( K^2 + K)
- Z4 Z" q- {- X$ g r! 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+ U, s1 b. O
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( v* v C" {: L& ?; Y = 3X + 3 ( K^2 + K)* P* i( _( b* `7 @6 _, [/ C3 H
= 3(X+ K^2 + K) which can be divided by 3
4 Q+ m+ k" b7 e" ]5 ~* r1 A
# b+ e# H4 U& M! y8 U$ v: y7 iConclusion: 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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