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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 e: Q/ \( a( |8 j$ l6 M& nProof: $ H% k0 u2 H6 {& l, E- Z3 ? O
Let n >1 be an integer 6 M+ F+ x3 M2 D" c& l+ n" }- _
Basis: (n=2)0 D; ^0 Y# E' R
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
* O+ h X# q3 B$ R8 o# g& ]2 v K^3 – K can by divided by 3.
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, W$ S" ` {8 z+ v5 f2 n" QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# w! Y; v! R: p/ b0 R1 }
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, r* L9 t' j y) P* @4 [
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 X: l. U. _8 L$ W
= K^3 + 3K^2 + 2K5 h8 ` f7 s3 s% q/ A; L% v
= ( K^3 – K) + ( 3K^2 + 3K)
8 [1 D, E7 a& ~1 n' U a& ] = ( K^3 – K) + 3 ( K^2 + K)
+ N! {" b3 W" }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
) W/ n7 F( c- p @2 u2 Q5 d# s1 GSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ K/ f! y5 L+ N5 C( J
= 3X + 3 ( K^2 + K)
! t7 s( s& ]% L# x* V = 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; |7 d9 K* T/ k3 f; u
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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