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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 |! ~+ k' O! gProof: }9 U7 b* V2 _2 y0 I
Let n >1 be an integer
! ?) B* P- [* D: O8 Q4 P0 ABasis: (n=2)% ?. ]8 b- a; k6 i; [
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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. n4 R5 J" Q% ~* i1 T/ \Induction Hypothesis: Let K >=2 be integers, support that( T0 p% F/ l% x" Y2 ]5 m
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$ x, H' r2 k5 I9 r
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' }4 Q) V& o$ u+ {# JThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% c6 R! H: t2 P. @' u
= K^3 + 3K^2 + 2K+ t- v! w: Q9 R2 D# ^) C8 J7 }0 _
= ( K^3 – K) + ( 3K^2 + 3K)
6 |( _/ m# p7 l1 W" n5 m5 q' [, ]( c = ( K^3 – K) + 3 ( K^2 + K)
$ z2 N4 G) t5 \' v9 \ Wby 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
8 G$ @% V* O& p, ?, T- OSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ G, B1 h7 R$ K = 3X + 3 ( K^2 + K)% a0 @ ]. e4 E/ T) p3 I
= 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 >1.4 \7 y: m* ^& q/ n
8 A$ D2 H( @' z9 W! R9 j; \: D[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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