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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)1 j1 C5 t/ E8 V& \
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Proof: 0 X. J% I# k! s! c! L- Z9 ^
Let n >1 be an integer ' [0 s7 g7 u( S) W4 b! q
Basis: (n=2)
( I' R) l" x' ^ E1 M3 p/ W8 p 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 {5 _) ]! Q/ `, q$ ~( K6 Y; \
$ f& B e i$ I5 bInduction Hypothesis: Let K >=2 be integers, support that
/ l- {( V7 h5 s2 V8 k K^3 – K can by divided by 3.7 S" U% G6 [, P: |$ v
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3- n% o$ X+ |; `: m- Z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
/ s) y) T4 d# c3 f* ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: r1 v* P% [3 N; K = K^3 + 3K^2 + 2K/ X8 i% n7 ]5 S: j1 c8 c- i
= ( K^3 – K) + ( 3K^2 + 3K)
7 k# j" h/ J! n2 |' N! H: K, e m = ( K^3 – K) + 3 ( K^2 + K)2 _5 K! d. Q% w) X) 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
& ` |' K9 m( J+ u- [4 m. vSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) D' S/ \5 ?7 E N, L5 I
= 3X + 3 ( K^2 + K)
E5 q2 i/ f. ]6 q$ c" H9 X = 3(X+ K^2 + K) which can be divided by 3. F+ G) w4 ?" N9 T
! G8 z! [ b' YConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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8 `5 c; _7 S' s+ D3 {[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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