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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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( f J' L5 G" @# h. d3 K! ^Proof: : X% w: Z/ ~9 ~# X% Y; S
Let n >1 be an integer 9 k$ |2 @; @' q
Basis: (n=2)' \0 t) w* n0 v% n! Z% E! a
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 c7 C q9 _) h* n
1 J& h6 `7 U" X" U& q+ D" |) F/ }Induction Hypothesis: Let K >=2 be integers, support that
! }* V: G) B& ~+ s K^3 – K can by divided by 3.7 k! F9 P8 o/ o U6 c- L! I- [
1 Y$ T* n [. B: E" h5 o% ?/ tNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) |/ y p) x. p8 I7 U/ S
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ r$ j( c0 v: Q
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' f' k8 d7 {. e+ {$ B" }
= K^3 + 3K^2 + 2K
- C( F, C3 j/ V! l( I& Z" U. { = ( K^3 – K) + ( 3K^2 + 3K)7 ]/ a0 B% @* ^
= ( K^3 – K) + 3 ( K^2 + K)
" l4 H. W2 V2 y. F" }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
0 t! w1 Z9 ], j( w6 e6 e5 o2 M+ uSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' h6 S$ {9 a) U ] _* q
= 3X + 3 ( K^2 + K)
5 J! P: u# d. E, {- R# w4 } = 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 >1.6 N' A, R/ Z) Q8 w+ C$ T0 Y
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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