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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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. p- ? M' \& [5 F" c! mProof: ) C" b. S8 | c6 Z6 L
Let n >1 be an integer . P& _1 M6 H1 G( I! |) C; N2 H
Basis: (n=2)
U# H" G& z: Z" U' ]6 E/ W 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 |2 R) D2 l* x& G; c. q
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Induction Hypothesis: Let K >=2 be integers, support that
; b! q& Y' w: X. p* I9 H" B5 n 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# B+ O9 m3 b+ n9 j9 l
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 }1 K0 ~# X7 `5 F9 E$ jThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
0 F1 f- K) h- ~0 m" C2 X+ J9 o = K^3 + 3K^2 + 2K$ E4 E0 F5 o2 |* R' o
= ( K^3 – K) + ( 3K^2 + 3K)* _: }, S G( e5 L# x$ D( G
= ( K^3 – K) + 3 ( K^2 + K)
2 O* O/ c- M V$ ?" Cby 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
a& x5 @# N# X/ d! } dSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 B& Z' L2 S# a F, `
= 3X + 3 ( K^2 + K)
?, Y; W* j9 q( l& x4 @ = 3(X+ K^2 + K) which can be divided by 37 N6 D! R- g6 V
O7 @: |( D' X" cConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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