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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)$ f# r% ~9 j1 Y: d3 w& g
3 T0 ~( w7 d' _/ j& k9 hProof: 7 c1 z: E- m; ^8 M% |$ A& K
Let n >1 be an integer & s" J2 f5 I. u* f; u
Basis: (n=2)' r, U0 q1 g3 X5 C/ s. Z- o
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
$ }% V# }2 ?* b; }4 n D6 t9 h1 V
1 ?; a# P+ r' Y' g7 vInduction Hypothesis: Let K >=2 be integers, support that) D# L( ]( \/ p3 [& D
K^3 – K can by divided by 3.
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. u# A; w: v7 I' j ENow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 ^5 Y) s* ~8 U1 T- F. Isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. r9 F) i' v ]' @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ z/ l* \+ l5 k' R/ [) D = K^3 + 3K^2 + 2K
1 Q- c# \" d, x, A = ( K^3 – K) + ( 3K^2 + 3K)
7 H3 N% @& L3 p6 W! p = ( K^3 – K) + 3 ( K^2 + K)
! G3 y( y2 Y1 Wby 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$ g$ `7 t8 I, s
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" A" C9 V" a, T6 L = 3X + 3 ( K^2 + K)
. Y4 f3 A+ @% G6 @ = 3(X+ K^2 + K) which can be divided by 37 N* N, C4 m7 w5 v; c- ]
/ G- K' ?8 N8 M2 bConclusion: 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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