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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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8 C: {+ K* k7 T, e2 Y, `* U3 u yProof: 3 [* G A" T& f$ D% t2 o
Let n >1 be an integer
0 L! c6 h% ?" M" G5 b4 ^% Y/ Z: c# D2 qBasis: (n=2)
* B9 W2 ]) d, E. y8 g 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 35 W, |6 o5 P6 D6 f& R0 L S* z
V: i# W5 _' T7 d9 CInduction Hypothesis: Let K >=2 be integers, support that: K. v* C5 l/ ^* D4 C+ K
K^3 – K can by divided by 3.8 H+ j, m# G* J
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
K2 o* z. A- A- l/ D2 G$ w' Zsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: ^# a* A# k% B$ M/ T5 s* b. i
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ ?! j4 W, T) j = K^3 + 3K^2 + 2K
; X5 k7 O) K% m) j3 L = ( K^3 – K) + ( 3K^2 + 3K)
$ m1 ~! h) }9 \0 O = ( K^3 – K) + 3 ( K^2 + K)
6 l. c2 X: C3 C. }9 H. oby 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
1 V8 I6 l6 q2 U0 w5 h7 USo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 K0 _+ Y5 ~$ [' J+ H) l = 3X + 3 ( K^2 + K). a7 ?% R* x. b0 b' }* U9 ~$ _
= 3(X+ K^2 + K) which can be divided by 3
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9 J9 H9 |7 U X$ |' tConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1." z0 B. K" k9 {0 r, Y% D* f
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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