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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 f. i9 S0 i3 E# ` x
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Proof: v. G( K n' J! b/ X$ _ O* N6 n
Let n >1 be an integer
& b$ C3 {; G) e+ V& C/ uBasis: (n=2)
4 n& v% q1 X9 p5 o 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 a6 q8 K$ ]4 \ X
: A2 c- F8 G# K/ H$ z0 v% m
Induction Hypothesis: Let K >=2 be integers, support that
$ `- v1 {2 K. v2 d, f5 [' b# c K^3 – K can by divided by 3.- p, q/ \" E" f8 Z4 k, Q/ z
4 W0 V/ r" q9 DNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3% K" f9 {! H5 H! S1 N
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ d& A& T4 |) r0 o: S4 ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! e! z5 R/ ~' W! b
= K^3 + 3K^2 + 2K, A" I3 m& w9 f- X
= ( K^3 – K) + ( 3K^2 + 3K)2 ?1 u- s$ u7 j
= ( K^3 – K) + 3 ( K^2 + K)2 V% c0 P( a0 c4 P$ J1 N6 ~
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
3 f+ x6 T; O2 ?3 `( N/ JSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% K9 Z2 b6 C) @1 ]
= 3X + 3 ( K^2 + K)
( c* F0 x% u; w- O, g = 3(X+ K^2 + K) which can be divided by 3
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1 [, r: I' }- h' vConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; \+ o' l- R$ {$ V
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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