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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:
; v6 P9 z i/ I! B) b$ ~& P7 ELet n >1 be an integer
/ K0 X9 x9 ?/ u. NBasis: (n=2)/ I) k7 W8 I. o- u
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
. D+ I1 i7 d9 S; D& ` K^3 – K can by divided by 3.' B, e7 T4 L- W/ M1 C% R- P( r
+ q" C. \2 A: h8 o- O" N( sNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. r+ I/ E7 {: fsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem @! U8 I- r2 Y! C% g
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: B; o; g* ]$ H& u% ?2 V = K^3 + 3K^2 + 2K
7 P6 B+ o$ U7 |2 J5 N5 H = ( K^3 – K) + ( 3K^2 + 3K)
: ~8 _7 l8 v# m# W3 d1 c! C# X9 y = ( K^3 – K) + 3 ( K^2 + K)$ S r# S/ O- B
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
% W3 v8 M( h: t/ U, z3 _' NSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! o) D% e# y# t8 u5 g W = 3X + 3 ( K^2 + K)
% m' ]- I. L, z. X& d5 a = 3(X+ K^2 + K) which can be divided by 3$ ^- [$ W; E' i5 R
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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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