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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)2 Z6 ?/ _8 V y; _/ K" l
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Proof: % O# c" q. P; }2 v7 u3 T$ @( P
Let n >1 be an integer 0 i7 O7 j8 t" `" u( K
Basis: (n=2)4 L4 b0 j+ d0 g
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) n" a( q1 ~* ~3 X* r/ E
& J! |, q$ ^6 n3 B
Induction Hypothesis: Let K >=2 be integers, support that7 V$ ~/ _' r# w9 k# D% r& y
K^3 – K can by divided by 3.& o0 e$ W# J( P, Y3 A) ~$ j c
5 w8 g, H9 L( e" U3 [$ KNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
; M% b1 [. o8 l% csince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 \. k% U7 _! z' _$ t9 zThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ C9 y" e7 b0 W# c% E
= K^3 + 3K^2 + 2K: v7 d$ P( l8 k4 k
= ( K^3 – K) + ( 3K^2 + 3K)" u" j! _1 j$ i
= ( K^3 – K) + 3 ( K^2 + K)
1 B& e9 ~# Q2 S0 g) Dby 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>07 K; ?! e5 G0 M. S
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 Z( v( C$ V- w1 s3 W. O/ u6 V = 3X + 3 ( K^2 + K)
" w% X" f5 m' ?7 i# R4 t% m = 3(X+ K^2 + K) which can be divided by 3
9 D* \0 c9 x- ]
3 V5 F/ [) G$ p4 _/ t% mConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.: K) W/ G( H4 H, M0 C
( }) Y6 V2 q/ n[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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