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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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Proof:
0 L# E" G/ B# H+ eLet n >1 be an integer
, B5 F4 @* ^+ {8 L) C, s+ P. z- ~Basis: (n=2)
( `6 U& _; K& p7 N& g 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 n7 a/ n* q: ~1 a: R- |
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Induction Hypothesis: Let K >=2 be integers, support that4 {3 }) e7 r4 {# U% b
K^3 – K can by divided by 3.# s4 h" k! P, G( P
7 u/ g/ ]2 X2 v0 E& jNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3. S7 w$ G& o0 i, v6 Q
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
4 _: ~2 h) _8 J f* n; lThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 i/ ~4 E6 w ~ = K^3 + 3K^2 + 2K
; y. D7 b6 Y* y8 `' p+ I' r = ( K^3 – K) + ( 3K^2 + 3K)- P* ]. X' u" k7 n
= ( K^3 – K) + 3 ( K^2 + K)
4 o' j) A% w1 S2 K; L6 C' Eby 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
5 ~9 \( x9 g5 r: r& I7 r! cSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* Y5 s; n4 D. h/ b$ R5 h = 3X + 3 ( K^2 + K)
$ _3 P8 Z |& Y2 S- c = 3(X+ K^2 + K) which can be divided by 3
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& P: S6 c- D- ~5 E; u: ~+ v. I! F) }Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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% k' Y! p* V h. l0 I9 K" i, E- ^0 S[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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