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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), w& K+ x5 F1 u2 _0 |
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Proof: " b" b* n) j. k) T
Let n >1 be an integer
5 m- Q) X8 f/ m9 m8 r) [Basis: (n=2)
% k x; P/ I- \$ _0 e 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 36 |( d5 U2 o3 c, h B2 M
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Induction Hypothesis: Let K >=2 be integers, support that
3 j" J- G' c* W4 c K^3 – K can by divided by 3.
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8 I, E* j# ?9 _3 I8 y2 D! yNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- O0 B! F% {2 r7 l& Hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# M3 [1 v+ P V' S7 b f1 W: vThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. P9 @3 V3 v+ N% _5 M3 S = K^3 + 3K^2 + 2K0 l' A' x7 O1 j% _1 m. H
= ( K^3 – K) + ( 3K^2 + 3K)0 j h0 c5 y6 z8 E
= ( K^3 – K) + 3 ( K^2 + K)
+ D2 c b' B! T" m: Tby 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; w6 S* I! u8 l' B) f, o4 }
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. ]( y: M; u4 Z3 Y4 X = 3X + 3 ( K^2 + K)
1 N* h. V$ v+ h4 y = 3(X+ K^2 + K) which can be divided by 3
* B- K: I9 t5 r/ ^3 y8 F8 U1 X! A2 G3 W" k
Conclusion: 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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