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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:
l$ W( d7 \0 _# y; x) [Let n >1 be an integer 7 C6 A. p5 R. V
Basis: (n=2)
7 R! F# F, z: \& K! o% Y! G& ~( T 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
6 b# p* D1 p: b) \' f8 s0 ?+ B 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 35 D7 Y7 u5 @2 E# h' Y6 c
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; ~' d9 c! m3 x# @8 uThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% P, v% x- i" H/ k' E% Q = K^3 + 3K^2 + 2K
: t" E: s( x( j! G: F = ( K^3 – K) + ( 3K^2 + 3K)
5 }$ L6 ^: s1 ]8 m = ( K^3 – K) + 3 ( K^2 + K)
! c7 R/ i" }- _+ e! H# y- vby 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>08 H7 x. l" U* U" g3 V- d
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ H5 f% Q. x0 _2 ] g% G7 }3 y7 t = 3X + 3 ( K^2 + K)
1 o5 l" J& o* }# }. n = 3(X+ K^2 + K) which can be divided by 3
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: S7 Z. H2 ^. q* n- D6 P' k" ^Conclusion: 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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