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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:
4 m% R$ T4 v: b2 M6 S3 g I* |5 f4 fLet n >1 be an integer
( \" g/ `6 \5 Z1 H' w0 r9 E l0 LBasis: (n=2)
! x7 I2 F" |) S6 H* C; @7 v# G 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 that3 z" t) a+ }% o; s" k1 N: V* H' l+ z
K^3 – K can by divided by 3.+ w P9 _/ \" T5 E `- c
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ a/ ^2 d" {, G7 H. i' psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
9 s7 n2 F; n4 N+ d5 E( l6 jThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 _9 ?/ X# y& g. _
= K^3 + 3K^2 + 2K
5 V" s- |; M- Q P/ M. p, ] = ( K^3 – K) + ( 3K^2 + 3K)
! C' \2 t* R$ j, _* } = ( K^3 – K) + 3 ( K^2 + K)
8 ^" {2 ?( t3 f7 {! d$ xby 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>04 o8 F5 U6 V5 L8 P) S6 Y& p! \& j, Q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* k3 d7 j ]& ^/ h, \ = 3X + 3 ( K^2 + K)
; g) Q9 o4 p/ c- Y* M = 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.+ w( X5 j7 i' P- O5 {) E' ]
0 d. j2 T; U/ P# B+ u[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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