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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). `& k( t) @; I- J! X: U
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Proof:
, F0 n/ k) E" d+ VLet n >1 be an integer . m0 \; Z1 I7 S1 `
Basis: (n=2)- _; _" @( ^9 t4 i% n
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 n% Q; C' n7 d0 }
, ]0 O! q4 x) O5 J. GInduction Hypothesis: Let K >=2 be integers, support that5 t( o& N# ^' l& o% \$ Z# @$ {- R
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
2 @- I4 g6 g/ E/ N) g' i0 j& nsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! k5 E: h6 r* a4 x
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 d" S( J; f8 q0 |$ C, y( i+ l = K^3 + 3K^2 + 2K g' b/ ~1 f/ z' S5 H3 w
= ( K^3 – K) + ( 3K^2 + 3K) v4 t- J$ j" E
= ( K^3 – K) + 3 ( K^2 + K)2 G* d1 M4 J; ^' o
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* e- k9 k ~, L$ W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) ?3 S& b, [0 J+ x: a( [' A# q8 w4 g
= 3X + 3 ( K^2 + K)
{- f3 t$ Y1 Y$ ?+ f = 3(X+ K^2 + K) which can be divided by 3/ ~! t% ^' z' r& g1 E" l
, M$ V" d& n2 t
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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