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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)
) V H+ J, D5 y7 J1 X& c9 G" _; n, v
Proof:
) ]5 @4 {3 V4 y' S @! VLet n >1 be an integer
- ]3 a* C) W$ A C4 uBasis: (n=2)
" Y- Y; E- k5 d7 c% v 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, Z* r: ~' U1 e) n# x* |! J1 J q
2 Z+ q+ R. [5 x3 \' d) W! q; C
Induction Hypothesis: Let K >=2 be integers, support that
2 n% F, ], I! _& m9 V K^3 – K can by divided by 3.
) T; I. C! M% A0 l5 z2 W$ {) {: h9 V/ d1 [/ s0 \! i) k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
2 F6 G. y1 l) {6 usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 w2 K8 D) ^: @$ K* W; j
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ g2 N4 R5 s( L3 a8 Q4 c
= K^3 + 3K^2 + 2K
2 }3 } ]$ E$ V7 I- [3 O1 Q = ( K^3 – K) + ( 3K^2 + 3K): x9 Q Y9 p) G; M3 O* e+ V4 `
= ( K^3 – K) + 3 ( K^2 + K), }1 R: o {$ k2 }* h" n
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& P0 j* V0 n( N6 G7 n+ R3 a
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! A0 y6 L! }- [: ] = 3X + 3 ( K^2 + K)& C3 L& |# B" L2 i9 @% w0 g
= 3(X+ K^2 + K) which can be divided by 3
0 i4 C5 A: o: z: h
/ y* f* u% f/ w' E3 D* t1 s" dConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.$ u2 ?3 B0 S4 E1 h+ j
s) X: s' `: s4 e' Z2 c[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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