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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:
: x* e& O+ T* o( k2 F8 K1 wLet n >1 be an integer - Q+ a$ C/ b+ r; a8 p( u
Basis: (n=2)
$ ^) {3 i) u& ~+ u* w! s( i 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 36 b( `9 a& e4 K _, S
# m0 p) e+ P: s3 {+ n
Induction Hypothesis: Let K >=2 be integers, support that3 L# x3 i" ^) O0 K3 O
K^3 – K can by divided by 3./ P; Y0 n/ a1 c! V' c. i" F
. {/ }4 J$ V+ T3 C8 R; oNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 S1 u$ D# \/ ~9 ?since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# S2 c' o9 f+ g+ ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 {! X( l2 B/ E: [
= K^3 + 3K^2 + 2K
; ~/ j6 e# k7 T- y* l' ]# s9 s% C = ( K^3 – K) + ( 3K^2 + 3K)7 ~8 V$ u. ^; J7 t9 ]
= ( K^3 – K) + 3 ( K^2 + K)
' M! G* a. H+ B) F. W6 E3 yby 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, T! K6 l f) [ T( u# |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# O& h" m: b5 T9 P = 3X + 3 ( K^2 + K)
- g/ }% w. R0 b = 3(X+ K^2 + K) which can be divided by 3& s* {; w, x2 t' _0 }- w4 }
. k9 S+ ], N0 {- Y- OConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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3 `7 `' {- A5 [8 ]3 C[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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