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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)4 o7 B7 G# a/ C
# W! @# q% Q J% x0 Q9 P4 mProof:
8 R8 D' w) Z$ {& E: M# U1 WLet n >1 be an integer
! ~0 P* ~& G/ h. \8 ]Basis: (n=2)- m* `3 s: }: P) r1 Q; }/ ^# u
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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( A9 P; t6 n; wInduction Hypothesis: Let K >=2 be integers, support that
/ W: C% k: I& X# V! U3 [ K^3 – K can by divided by 3.
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1 \. z% M# V6 o- X! L. J+ YNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# X! T8 f+ I3 e5 @1 b; w- W
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 E4 W3 ~! V8 Q" d; a# z: Q% u' c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ t6 C# Q( z/ }. c+ D
= K^3 + 3K^2 + 2K# I. S% a0 E7 `8 `- O
= ( K^3 – K) + ( 3K^2 + 3K)4 e0 u4 Z* E+ R0 U' `+ x" ^9 T
= ( K^3 – K) + 3 ( K^2 + K)1 C- J$ Y7 |7 c# m# Y
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- n i1 ]3 a5 `5 o! p0 ^7 f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 g) z- r; s {; N% {
= 3X + 3 ( K^2 + K)2 l0 ^6 q1 o9 |' J4 |1 x4 J
= 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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' `/ Y% d1 W2 d7 M8 b/ f[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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