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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: 6 S" w5 Y" q5 Z |8 `4 x
Let n >1 be an integer
; }8 h1 c# [; d2 d i! GBasis: (n=2)0 p0 E3 i4 s2 k* I
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 that1 [! g, a- O* F: V5 i) v% f
K^3 – K can by divided by 3.
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9 {3 o( F, `) |: w4 E/ V( _Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3& k0 @7 _- x8 ?
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( o, Y1 F( I3 c# ~, E* Y) NThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% n, E" s$ {" l; ~- | o) j- r) b" w = K^3 + 3K^2 + 2K$ T X5 ~4 D+ _2 o* c& B$ |
= ( K^3 – K) + ( 3K^2 + 3K)1 H2 P: U; _0 `8 I( O- r/ d5 g2 `$ \
= ( K^3 – K) + 3 ( K^2 + K)
$ U, R5 h5 ?+ b8 {+ ~8 aby 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
9 _+ [% J! e% ^6 X8 Y! FSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) E$ f( F" ?/ [. {0 Z# R) S
= 3X + 3 ( K^2 + K)
1 I' F) I5 P: S4 {; e = 3(X+ K^2 + K) which can be divided by 3# H* J& B2 O. c2 t. \1 Q# s
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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( T0 g" [; g9 P) i: p# n1 ?: `5 `[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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