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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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4 M3 a& n' `3 ?1 K+ U" q! l" c/ PProof: ! l* \1 R9 R, Q
Let n >1 be an integer 6 t9 S3 j) c, b3 K- D
Basis: (n=2)8 J3 P. w9 `4 ~1 b1 ?4 D
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( b+ M C& h/ z4 y0 @9 S! J/ C
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Induction Hypothesis: Let K >=2 be integers, support that
/ {* {9 U0 m' }* X& y K^3 – K can by divided by 3.. c; Y1 q, B; J. l
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ _) s( k7 V) Y: I! Y) D& Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& l& g9 O7 K, nThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; a/ T% H5 Y6 k: Q7 ]3 }7 b = K^3 + 3K^2 + 2K U3 a) x% k/ m# g. y1 Y1 c
= ( K^3 – K) + ( 3K^2 + 3K)
5 P, l: C9 C2 \% U = ( K^3 – K) + 3 ( K^2 + K)
: I$ W! r" _- M1 G3 B$ Dby 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>04 r1 U; b) }- ~1 `+ _4 q" z
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& v* a3 O& s4 q4 |* e8 U1 k8 D+ ^ = 3X + 3 ( K^2 + K)% T7 p. L8 h( ^" I: g! l1 c
= 3(X+ K^2 + K) which can be divided by 33 W2 [0 p( H: C( f+ J. M
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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