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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: $ b& X# G9 U+ h% j
Let n >1 be an integer 6 b- d5 j0 g2 T( v
Basis: (n=2)
/ A- G% Q# Y* b9 h4 F( k% m7 x 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' P/ b5 W3 E2 D- I, y( J
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Induction Hypothesis: Let K >=2 be integers, support that
! Q; S1 k' C2 ~7 b K^3 – K can by divided by 3.
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3 C' r! e \ o0 ~& o$ u/ p0 {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 q4 e0 `1 j asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 }2 {4 O. l0 b2 g( j k4 l& VThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
# R( \# W8 m& N7 a. r$ ]9 s, d = K^3 + 3K^2 + 2K
8 Z1 z5 `0 _' T2 s( s+ ] = ( K^3 – K) + ( 3K^2 + 3K)
: i4 A* F- O" Z; T6 V- F = ( K^3 – K) + 3 ( K^2 + K)! x4 R0 q3 M/ |
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
$ s7 D9 a* X+ V; [' C5 zSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 |; S( D# z/ c# x = 3X + 3 ( K^2 + K)
6 i6 T2 N2 l: A" X6 R = 3(X+ K^2 + K) which can be divided by 30 g) g" G: H, B5 o2 r
" {4 ]/ `6 h5 M: GConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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' ~& U4 u7 r+ k* G0 U2 A' H2 ?7 Z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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