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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)6 Z6 F, r9 g* P9 ]( C& j! e/ H! u
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Proof: % x; X/ \& E$ t8 l/ X. D ^# F
Let n >1 be an integer
& n8 }' ~7 @" \0 k0 h2 ZBasis: (n=2)3 I8 J4 c4 V& U3 X* Q ^
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+ S3 w4 V5 ^6 o6 Y# N
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) o! |7 }# p0 G/ L( M* |
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, J* j: @2 }8 | r& H& a
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ t, h2 q; U8 P/ X4 o = K^3 + 3K^2 + 2K
. I% t/ t" n2 Y' q2 b$ ^ = ( K^3 – K) + ( 3K^2 + 3K)
+ V: X- B* @, U) |/ @! a = ( K^3 – K) + 3 ( K^2 + K)2 k: m0 ~6 ^. S/ L
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
/ E6 r, k& Z8 b5 s) N7 C wSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: V1 M& ~( M5 o = 3X + 3 ( K^2 + K)# j: q6 k6 c- J, M* o
= 3(X+ K^2 + K) which can be divided by 3, U" ]7 e, C* M
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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