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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)) H/ a* {( Q* G
$ s2 r1 w$ y# k) x- mProof: & x% {7 n0 k* U
Let n >1 be an integer * Z# l# V/ D! d; Q `( R
Basis: (n=2)
" R3 F9 B7 |/ ]- i+ I 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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, M: @5 z" ]5 j0 h/ O0 f/ mInduction Hypothesis: Let K >=2 be integers, support that
\. [4 I4 m8 u S K K^3 – K can by divided by 3.) G9 x8 [& J8 ?+ Q# M$ y
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
2 x$ |- j& s; D( o! Msince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 l% T7 I- E' p+ u! @
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. V1 v% c& j. \/ D2 ? = K^3 + 3K^2 + 2K, g& \( S- S u2 i' ~
= ( K^3 – K) + ( 3K^2 + 3K)
/ k9 p! i/ Z) h( ~$ Y# f = ( K^3 – K) + 3 ( K^2 + K)) N5 m0 q& D+ ~* U( f
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
) ?7 F, {- g' X9 @. ^0 XSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 f, x* r6 G; G* d3 E = 3X + 3 ( K^2 + K)8 u0 x3 a- {% R0 N" r4 S {9 E
= 3(X+ K^2 + K) which can be divided by 3( U6 O. G4 n: _! Y$ A" q+ [+ Q
* ]6 E( T: x/ H. I4 f$ U6 v3 J) b
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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/ h! A% g. A3 V# M% [( g1 x2 Z+ Y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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