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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)% Q- B+ B+ G% F% h% S3 N6 e n
* D* i) E5 t- J' ZProof: * T/ h6 |0 d& _9 y+ T/ n
Let n >1 be an integer - w: |. ?% L: A2 Y
Basis: (n=2)) k, q l% X7 l6 @- `
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
7 a5 k3 P( m( d# e6 n$ f! P5 i B2 W% i* |+ h. y' S8 o1 M. X. a
Induction Hypothesis: Let K >=2 be integers, support that
) t% l6 R/ J6 l w) b K^3 – K can by divided by 3.% R9 Q" N) L8 t2 H/ v
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 q& [# K6 E" Z" s0 c8 \since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 h8 ?8 Y0 G- z& S4 t
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 H9 O8 m, N3 x! m
= K^3 + 3K^2 + 2K+ e1 w7 }; k' U5 U; m. J' Q5 y
= ( K^3 – K) + ( 3K^2 + 3K)- H/ x9 c* V8 c, [) V( F" t" n1 n
= ( K^3 – K) + 3 ( K^2 + K), l4 m9 K# e3 b
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
* a( S7 j; Y) E5 P& R+ I6 | ?+ _So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& P8 d1 P8 Y: I) J5 ~) k+ [. E6 b
= 3X + 3 ( K^2 + K)
1 u& _2 Q' x" e: z$ f. P! ?7 G6 P = 3(X+ K^2 + K) which can be divided by 3
$ E6 d5 z4 a" ^; x4 @) T
- c8 Z& z% P6 vConclusion: 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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