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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 I& b/ v: ~) ]
$ |9 X3 O+ [4 N; hProof:
9 ?8 u5 o; M" q; c: E, x3 `Let n >1 be an integer
* z5 V8 C# G2 MBasis: (n=2)
& l( {, e/ ]8 I; `! K! Q1 ?) p4 a 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 M) ~3 ~ k( o1 ]
; p9 M2 ]3 a- v m0 n
Induction Hypothesis: Let K >=2 be integers, support that& D1 t0 a9 ^/ Q
K^3 – K can by divided by 3.0 b& l$ y" f6 O3 `2 g4 \7 v
p2 Q& r; m$ S9 M% CNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
; r+ U t% O/ q/ M4 qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, ?6 m' u' P# d; x KThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
r4 G, _( d: |; ` = K^3 + 3K^2 + 2K
4 p, g% g4 W; s = ( K^3 – K) + ( 3K^2 + 3K)
. ] t! p# g7 s( C = ( K^3 – K) + 3 ( K^2 + K)
$ a) `& d# `* k P3 |" e- Z1 Hby 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
: F5 J; m! w( e; WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; L$ h5 ~5 K, ^7 o* u = 3X + 3 ( K^2 + K)- m3 U" `+ `$ W% X% ]* D! T* ?
= 3(X+ K^2 + K) which can be divided by 3, G# O; E8 `$ T
1 T1 C* d7 }- |4 zConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
% K% a4 e% h/ V/ ~& E# y! i' n" I, Z% m7 ]
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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