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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)
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Proof: ) r4 Y% ]. ]! E6 W" D7 e
Let n >1 be an integer 8 l" `6 R: f& u0 ~! U4 M
Basis: (n=2)
8 U, s( U+ }# g1 q8 K$ \1 A1 Y0 { 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' j! Y# _+ \8 K1 U/ y: \% Y
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Induction Hypothesis: Let K >=2 be integers, support that) |3 S! e* A- [6 \
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& r, N; r8 ?; |5 H2 J& gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
7 _3 X0 W$ U' \8 r- U2 ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
! q% w$ {6 i2 V. B = K^3 + 3K^2 + 2K* _# ^) N! W( s$ d
= ( K^3 – K) + ( 3K^2 + 3K)3 T F: i8 {" n }9 V- \( S3 P* T
= ( K^3 – K) + 3 ( K^2 + K)( } d2 N8 |+ ?+ s# u
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* F: \5 ?8 t! E8 b) R
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! w. f% F* a& V4 {* T8 v# `0 f
= 3X + 3 ( K^2 + K)
3 n* w2 O3 y- Q1 ~ = 3(X+ K^2 + K) which can be divided by 3
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) [ d7 D6 a2 D' X. x! jConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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/ v1 y$ f' m4 f# e$ {; C. D- A[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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