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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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: } i4 u1 ]8 q" [7 j3 v" QProof:
* H7 Q) G* w" N3 \6 [Let n >1 be an integer 9 R3 V4 P& [3 D; S: o% |5 Z, U' g
Basis: (n=2)
7 D: z# F, i( H9 U$ v" p0 o 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( G& D# { a; Q
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Induction Hypothesis: Let K >=2 be integers, support that+ C; |# [) u& w/ L' N
K^3 – K can by divided by 3.1 K. p( `: J/ S8 G: s; p% Q
; F9 W2 f) N- u7 P9 }- a5 ~: }Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
" H& q# T( r( r' ^& a. M3 vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" E2 D2 X) @+ F5 W1 F% ?
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& `) W3 r: V6 c; Q3 F- L
= K^3 + 3K^2 + 2K
Z$ x# d: u, h1 D' ~5 [ = ( K^3 – K) + ( 3K^2 + 3K)% W* y k( R" E
= ( K^3 – K) + 3 ( K^2 + K)
3 @$ N! K* S7 Vby 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- P3 G; h* Y1 @, T% K( g- j8 b
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 _- o/ N5 ?, ~4 P1 `& T = 3X + 3 ( K^2 + K)
2 C6 O- j6 i1 A! F$ \ = 3(X+ K^2 + K) which can be divided by 33 i7 K u8 |+ J1 e
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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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