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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)# E4 F* M3 ~9 L, w
2 Z2 }, s9 D. e" TProof:
0 f' I' W' f H) X# y6 ELet n >1 be an integer , c; e8 ~" b) y4 J9 Q
Basis: (n=2)% f0 N7 g# c( X U3 B9 K+ H: Z4 z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) z4 Q# b h0 ^% c. ~' ~& J' x
B" I& M1 X! l6 k3 M8 ~# _
Induction Hypothesis: Let K >=2 be integers, support that
+ V" _: b2 S2 N. [ K^3 – K can by divided by 3.( `7 L% {, j, C7 f+ G
$ f6 C: [( F! m% r; SNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 @4 J8 y2 y6 u0 Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ s+ w- b6 p, C$ c2 n0 W6 {# m
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ P0 q7 h* Y& J K* `0 E = K^3 + 3K^2 + 2K- T* M8 w+ \: _# `1 L
= ( K^3 – K) + ( 3K^2 + 3K)
3 A0 H9 w$ I. a* s7 m' a = ( K^3 – K) + 3 ( K^2 + K)0 b- v: F) O: J Z) k* X# C( Y- |) D
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; P& x& h4 N& j8 W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' _+ w2 w$ ]# g p. d3 Z' Y! I* I = 3X + 3 ( K^2 + K)) X4 s# f. \6 I5 d' C
= 3(X+ K^2 + K) which can be divided by 3* { ]- z$ E2 I6 F9 |# T- X1 t
6 ~5 g6 n; P2 |7 J% ]8 gConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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) H3 j1 u- H5 T4 w[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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