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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)
( V" s/ L6 ?1 A& k% a+ I4 ~0 L. p/ U0 P' i$ z7 L
Proof: % w4 J9 Q" V& m3 N1 n
Let n >1 be an integer + \2 A/ h: F; _. V
Basis: (n=2)
5 j, V) @: @7 F& L 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
: `8 }1 G h2 f1 Z. {( w" V9 O. ?) s" Z3 a* m7 C! }& K4 t# L( k
Induction Hypothesis: Let K >=2 be integers, support that
8 y+ D/ [1 H8 e3 w& ^0 c K^3 – K can by divided by 3.: k. q9 u) R4 D* f7 f. x
8 J) `2 X2 P" ~: e: {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! c/ y! L( f. U! [5 K4 Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, e, s1 o+ A$ ^1 ]3 F" ?( mThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* s% O& \" V2 S( R = K^3 + 3K^2 + 2K s9 k2 p0 S+ y
= ( K^3 – K) + ( 3K^2 + 3K)" m9 ?9 k: ~* Y6 J6 r* j
= ( K^3 – K) + 3 ( K^2 + K)6 k' Q4 T" \( N
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
6 U' L# N: H) \. M, Z, jSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ |1 @8 N' T, \0 k) ] _. \$ j0 s
= 3X + 3 ( K^2 + K)# Y+ u' C6 X: c$ p- `
= 3(X+ K^2 + K) which can be divided by 38 v0 }1 f' ~2 t
. Q0 x8 |6 }( S4 T0 G- NConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 K V" P! i: P+ c
! H' U+ f7 b1 n+ [. B1 J& ]2 J9 x6 {
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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