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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)
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! e# {$ }% C1 oProof: 3 M, A( O. } k- U$ L0 c! Q
Let n >1 be an integer % L; u0 E; p2 J s* T
Basis: (n=2)5 A. w' f, }) Y5 d
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 that0 G; r; d, V6 w; g9 F
K^3 – K can by divided by 3.
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( `: T" I% j n/ XNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 [& O& c- U! o4 b, p9 |4 n# usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, u6 p9 _. q4 I" c; N+ L
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% z5 o+ w9 |( ^8 }/ n% p
= K^3 + 3K^2 + 2K
/ K$ s: [, w8 t! p, g0 w = ( K^3 – K) + ( 3K^2 + 3K)
0 V, v* I: T8 Y0 Y" R& O = ( K^3 – K) + 3 ( K^2 + K)
8 N( g, x: Y V% x: dby 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>03 ]( Y2 \- y. {2 q% u% s
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)6 B" K2 _: `: k/ r
= 3X + 3 ( K^2 + K)0 x: D9 K1 M$ E" `# P
= 3(X+ K^2 + K) which can be divided by 31 G% b9 h u% C' V: p$ l! Y
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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! O& C; R8 n- f! i[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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