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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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: |6 U& d; v0 q2 D( qProof: $ d, p2 i) c: E/ B5 a; f
Let n >1 be an integer * T8 d4 P8 Q/ v. j4 C
Basis: (n=2)$ @3 P9 @* k c3 u6 J+ ^+ ? a
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
8 F5 \0 V7 ~% K+ |; P% B& h 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: L; C$ \4 [$ L5 a! u
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, e* D. q) s5 F4 L
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 \) k* T# ~6 c/ H, N' \* q
= K^3 + 3K^2 + 2K
3 _2 D1 I; f6 U2 `$ `# O& @# E = ( K^3 – K) + ( 3K^2 + 3K)6 y2 v3 }/ J- A8 z' g! F" [" M- i
= ( K^3 – K) + 3 ( K^2 + K)
' A: S; h2 R# `$ D1 tby 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
; q" t* W' D: E s* SSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* ?- P$ \9 G3 u0 D6 b* z+ s0 } = 3X + 3 ( K^2 + K)
3 i9 D) S$ Z0 v = 3(X+ K^2 + K) which can be divided by 3 m1 `- I! E+ q- O0 i6 l
7 c) S: G. y' I \1 j+ }* pConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 n4 s' W- \: Y# b! Q
9 x) J$ w' K9 h5 ~[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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