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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: " y: t9 l/ p1 n* o; W
Let n >1 be an integer ) Z, B" ]# O0 b
Basis: (n=2)3 B5 y8 d! }- [& ~2 x/ V3 v
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 that
- _1 B' @2 r0 ]& m6 U, j K^3 – K can by divided by 3.
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; u9 x; w1 M: l5 Z4 k: i" R* g1 ZNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' @, r- W: m/ o0 K, zsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
/ e0 F+ O/ U9 pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
) P1 P& E# Z: a, _) j7 l = K^3 + 3K^2 + 2K
y( {) }2 \* G$ w& ^# |7 w4 F) g = ( K^3 – K) + ( 3K^2 + 3K)
0 v# H/ b5 ~ e" A = ( K^3 – K) + 3 ( K^2 + K)5 ~! o0 I* G2 [/ @# i
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>0, z4 N$ v" d8 I7 }! K& I8 c5 F/ s
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 u v+ o9 o# J; q = 3X + 3 ( K^2 + K)& O. N8 Q* s4 c
= 3(X+ K^2 + K) which can be divided by 3
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9 s' J& \0 E$ `0 n4 i: f0 kConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.4 X+ G6 y8 F7 Y- }
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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