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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) r# E/ h- }. d" j& Y
3 R0 p5 j, @$ {
Proof: ) R: w+ c. `. y* y% k
Let n >1 be an integer
! R: u8 E" m5 H% EBasis: (n=2), x; Q3 s+ B+ \# M; C9 V1 j
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# i0 a# W" G7 i. W. X
% T. y. R# h$ B j# j
Induction Hypothesis: Let K >=2 be integers, support that
" U5 J( i9 z: T1 y6 [& Q; P, ~ K^3 – K can by divided by 3.
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7 r+ D/ @* u$ P! k* n$ N7 }3 F- ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3; K) S7 X8 K* j$ ?# w3 a% U6 P9 s* i
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: ?" x W7 H8 a- }; s5 WThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: _, w: T! }0 }7 C2 L, t = K^3 + 3K^2 + 2K: c. ?9 o& M, j8 i- ]3 j
= ( K^3 – K) + ( 3K^2 + 3K)2 i/ l3 z: g3 g1 a& V; A
= ( K^3 – K) + 3 ( K^2 + K)
7 F; H) w9 K5 u" Z4 e2 Iby 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
" n5 b, v+ v' A; |) zSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
T, a* @+ b: S- `2 C = 3X + 3 ( K^2 + K)# L/ Q# `% D. r$ c
= 3(X+ K^2 + K) which can be divided by 3# g: b! ^; }1 ^
" c7 N" ^& Z( c9 X$ d* ?# R( FConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% m. k0 v# v! [8 q
. _3 e# x6 }2 g/ Y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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