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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)& X+ s6 w$ b, B! L) _/ N p. U n
# g4 Z8 J8 Z$ N4 O$ [/ `+ Z* y+ MProof:
9 l9 o# I& Q1 u7 T% ~Let n >1 be an integer , t) F2 \& M/ K2 u/ f: K1 K
Basis: (n=2)
% k$ y" {$ K( u6 y, {% E2 | 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
7 m7 L3 g* f1 y% D7 {1 K$ x: l* q% r# D! Y
Induction Hypothesis: Let K >=2 be integers, support that
1 f; ~5 m4 }$ C+ F( p# J0 n: k K^3 – K can by divided by 3.8 O2 z! O+ u4 j) w! w6 H: {4 C A
s7 G* ^4 T4 kNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ W. H5 M: V' m. c5 e7 A) `# i
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! n) p4 L' M8 O7 D
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, E8 i6 G K3 g = K^3 + 3K^2 + 2K
+ Y) a$ I! C& g7 a0 q; k+ x+ k- Q = ( K^3 – K) + ( 3K^2 + 3K)
5 u3 Y) V" j$ l @2 U6 t = ( K^3 – K) + 3 ( K^2 + K)
2 _- K1 d4 }6 ~7 T# z% V" q4 }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. t+ ~, k7 U( m: c
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ v. a u- R/ V0 h9 t! F* {
= 3X + 3 ( K^2 + K)
$ s; } p+ G' _ e5 t = 3(X+ K^2 + K) which can be divided by 39 f: z& x6 U4 h9 S: \, r7 @, R
, u5 P( H( W6 k( z* z
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., W$ V( o& w6 d5 K
" u4 E9 d: N6 W8 k: B[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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