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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)% p* y# K4 p/ K+ A( |0 ~, [
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Proof: ) q; E: ], t. s, d2 ?4 O& g: q
Let n >1 be an integer
' z8 c5 Q% P- |' i% L5 @; J8 UBasis: (n=2)
, u0 f! j4 d6 \( [) \; W1 B7 w 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
& b# n0 @2 |6 e8 C# J# d) {! h
' P1 H0 D/ n' D3 v6 eInduction Hypothesis: Let K >=2 be integers, support that& m& R0 W+ O: G% L) h: |2 x7 Q
K^3 – K can by divided by 3.
# i6 K; s# R; u4 L
0 ^: g* f! l; H0 t6 K9 a2 QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 30 c* G; a3 K* W- Z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem2 p9 R' {& _ w& Y0 s% M7 E; ?
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 Q5 B' g+ r9 j& m
= K^3 + 3K^2 + 2K
% y! Q$ ^4 P& \- ~' N( m7 \ Q1 S% R = ( K^3 – K) + ( 3K^2 + 3K)
4 @- Z; _$ c+ ?8 O( ?2 I7 s) n = ( K^3 – K) + 3 ( K^2 + K)$ [/ n$ L' c( |- N# X$ H
by 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>02 Q$ l% G/ D$ N
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ `8 c9 c: E, s; ?* R- n = 3X + 3 ( K^2 + K)( {+ `# ^9 N P9 Q6 ~
= 3(X+ K^2 + K) which can be divided by 3
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/ {- S/ o% b- Q8 GConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 |1 P) N+ |1 W2 P& r
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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