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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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: g4 x# w1 l* J) @' ^Proof: 3 m& t, |* `+ ]- i
Let n >1 be an integer / k2 s: B. ^8 v8 b/ }
Basis: (n=2)
+ I. P& T$ w9 T+ O8 R0 E1 _% A' X 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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X" ^4 h3 G2 n# T j& k- }- ~Induction Hypothesis: Let K >=2 be integers, support that
+ M! c1 \5 o5 n8 B0 k K^3 – K can by divided by 3.
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3 i# M+ n( P, G& P! G! tNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ t9 S9 |- y; B, T' @% a# m. Y; Nsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! _- X: Z9 X( T, iThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ i. e' I5 w; M q+ o: A; J5 } = K^3 + 3K^2 + 2K
5 G9 |# V" E, ~ = ( K^3 – K) + ( 3K^2 + 3K)9 K% J4 I1 l# u" [% x7 u; L8 K
= ( K^3 – K) + 3 ( K^2 + K)6 B" U# Z) ]* B! d8 ]
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
( B2 h6 G& L( r, V+ z* ?% jSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K), X: t$ c9 D2 t7 h" F& D! \
= 3X + 3 ( K^2 + K)1 A! t9 P7 u' A6 p [! }7 _( y
= 3(X+ K^2 + K) which can be divided by 33 @$ l6 F+ Y" V1 p4 {
. `+ z2 e$ o* I$ f* O1 I% ^$ HConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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/ d4 r5 |: ~+ U/ O# a[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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