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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)7 A: O7 ~+ t c. C- s
+ b& |; ?6 V1 T7 h4 i" KProof: 6 J- U- l* Q. q
Let n >1 be an integer
- s* X! Z3 L: p9 R. uBasis: (n=2)
6 y1 i1 j& @! K0 S( R 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ t5 P- W* c+ c) j1 K; d( h
+ N R: l% w8 _6 ~6 c3 I' EInduction Hypothesis: Let K >=2 be integers, support that
. | U' L& }4 q, \9 [ K^3 – K can by divided by 3.9 D3 S Q3 {4 V1 p% H6 `0 G# e
7 E" X/ l0 c S2 u, p7 }1 i1 e ^
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ q. t/ S1 h2 D H% v# B, A1 I) vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
7 e7 [ u" X0 rThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- \1 x0 M8 Z7 w6 R" c& Y6 { = K^3 + 3K^2 + 2K6 m7 r! M% V5 k/ k
= ( K^3 – K) + ( 3K^2 + 3K)6 u+ f2 a0 n: a6 C) U$ K, m
= ( K^3 – K) + 3 ( K^2 + K)% t% X' U Z/ A4 u& r
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& m2 F4 @& b& V& h, F
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)1 O* l8 `) f0 Z" i
= 3X + 3 ( K^2 + K)
9 j( A0 }1 G/ G = 3(X+ K^2 + K) which can be divided by 3
! j! s! q9 n% S
0 I; e0 j9 e, u- b2 X. fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ B: E# i P9 s0 S3 w
# @, O; @; W, X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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