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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: ( l; s, l8 Y" e6 I7 d% f
Let n >1 be an integer
7 j% V& Z( b. m1 }: Y0 f* fBasis: (n=2), ]. {* _8 K; V/ b9 i9 `9 b$ G
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% [+ R9 M3 i+ x4 r6 e2 y. C% T
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Induction Hypothesis: Let K >=2 be integers, support that
$ d. R: e* W) d7 O) e* O K^3 – K can by divided by 3.
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8 N% o6 m2 N) uNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 34 ] X9 g/ k5 K; D1 d! \
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem$ [ _! `: g* K# F& K b1 I% L* e
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
5 Y4 C; @9 d$ m9 K% e/ P! N = K^3 + 3K^2 + 2K: i0 v! ~+ s1 f$ @# Z6 W
= ( K^3 – K) + ( 3K^2 + 3K)
! ?* ~- Q' u' \, o+ x9 } = ( K^3 – K) + 3 ( K^2 + K)
$ H w% |" ^8 ?5 ]0 oby 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
! e8 ^4 G; o7 N0 l* j9 rSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K), q' v8 H0 P. p! u6 f
= 3X + 3 ( K^2 + K)
0 D& {2 K' q$ ^6 W, z = 3(X+ K^2 + K) which can be divided by 3
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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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0 R5 d. f, m* h[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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