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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)/ d9 s5 B/ s: T( C9 T
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Proof: _. G0 ^' J! i) X l# v
Let n >1 be an integer
0 F. D$ ?- A/ z2 L& XBasis: (n=2)
: P) w# d( B5 h% o1 a 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# j' g. L% g& m
1 {% c3 c6 Z; f3 LInduction Hypothesis: Let K >=2 be integers, support that. c( h2 C6 n8 I" q1 ~
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
$ b4 s9 c! e2 l% y% o5 |3 {since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! Q# ~, B! F. z9 Y! n4 Q% v
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)) [/ ^8 L" z J9 m0 }
= K^3 + 3K^2 + 2K2 E2 n( f* M4 o
= ( K^3 – K) + ( 3K^2 + 3K)+ }7 [- S8 |3 f! F& I4 D( n' U+ x
= ( K^3 – K) + 3 ( K^2 + K)& H# Y$ D5 {7 j) a4 U4 h% j
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
h4 J$ n$ Y8 Z9 N: D+ GSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); A" l1 Q( {% P' |& v7 O$ O
= 3X + 3 ( K^2 + K)+ F. q: ~& _9 _( J" t0 Y
= 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.
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5 h+ Y! `( k- m0 o. @& [7 F: v$ _& Z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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