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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)4 L4 i1 }/ z. D% l" s) F
! o2 n" {0 P& X0 f) d8 L
Proof:
8 E2 ?; Q0 r" L5 d; i0 tLet n >1 be an integer $ w( e' U8 r9 ?2 A: ]" F
Basis: (n=2), l- \6 `; u2 E% j& ^ z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ T* V1 _* }" |) u. X
" p( _* p1 T# ~. K+ y0 U$ iInduction Hypothesis: Let K >=2 be integers, support that
2 V- d5 `# q$ L; e K^3 – K can by divided by 3.# T# K% j" a* o1 [6 r6 l7 e
0 Q3 m1 d' `: y0 g" z$ H% G/ U; XNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) \$ c7 d% l1 z, s
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 d* W" y, t3 @2 k) C% _3 m
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 w$ Z- K! G; `% a( }6 Q! `
= K^3 + 3K^2 + 2K" L7 [* h; D5 k4 f, n$ v
= ( K^3 – K) + ( 3K^2 + 3K)
v. {) _- \$ B7 l V; W = ( K^3 – K) + 3 ( K^2 + K)
8 y, D5 r/ \# ? d$ Z& h$ 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>06 H1 N* j1 L3 i% [6 c q, h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% b) N2 Z: u% [7 J# s
= 3X + 3 ( K^2 + K)
- b6 Q' p" V' j& [ = 3(X+ K^2 + K) which can be divided by 3
6 n+ r( C& K" }5 ^
& y4 Y' K+ b' O) S/ VConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 y7 s' I6 m+ a. C( c: P
2 ?2 G# G" h6 j. w+ A[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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