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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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7 ?- v- h) q* K( s. ?9 _6 @Proof:
. h2 F+ S) L! J3 @0 P. M4 \Let n >1 be an integer
c" E( T. v/ V" O& hBasis: (n=2)
( B1 K! a' S' C6 b 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" v* y: Y- ~; ]! @; G
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Induction Hypothesis: Let K >=2 be integers, support that
( v( c" C+ y: B 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
5 v/ l" s& n9 Y/ u) l% R4 U, [since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
* E' T% e1 ~: G9 aThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 \- K7 R5 _% r$ z. J = K^3 + 3K^2 + 2K
( f$ M8 w% H+ K = ( K^3 – K) + ( 3K^2 + 3K)
' p! G6 ~# B* b4 a P( K2 z = ( K^3 – K) + 3 ( K^2 + K) L$ [5 f3 N$ T. Q% u. r
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
8 i0 O4 L/ C/ X! X1 I' x' zSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)1 K7 B# [4 G" h! E4 J: G3 p" S
= 3X + 3 ( K^2 + K) h5 D R3 @0 |
= 3(X+ K^2 + K) which can be divided by 31 e; D, H" l/ b- ^9 L- D0 L
4 ~4 @/ Q7 w: r9 w% YConclusion: 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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