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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:
/ s3 R2 i: K# z KLet n >1 be an integer
: c' y+ N7 w" z `; r- B, s' \Basis: (n=2)
5 z7 F8 R: y. j. y0 T 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 a4 S; ?. s) t r0 ]& r
' k: ?! E% @: Q7 k9 x8 _* TInduction Hypothesis: Let K >=2 be integers, support that' w V5 I* ?. V( w
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
4 p* ^1 I/ K4 |) i/ isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ G3 `; a+ v4 q* n
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
& o9 @' d# p: N7 x7 E = K^3 + 3K^2 + 2K
# ?4 m9 u# ]5 P0 r! b2 I6 g = ( K^3 – K) + ( 3K^2 + 3K)
; @$ X3 D9 I: D1 r( _ = ( K^3 – K) + 3 ( K^2 + K): N6 S' c' x& m% e! p
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 W* K& H( ^8 s# A3 `# N6 {9 ~
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)- c1 ]( e2 g- g K
= 3X + 3 ( K^2 + K)
+ s J5 [6 c' D& o = 3(X+ K^2 + K) which can be divided by 3
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# X, {# y B7 u( W) W% |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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