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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 C% P* e+ P* c1 Y7 z1 |/ ^ R' h
( x3 Z0 Q* m9 b8 s+ ?: t, N! KProof:
, _ w0 w( i! X# D% P) HLet n >1 be an integer + S4 ^% {& d7 M0 K k9 l
Basis: (n=2)2 W+ {4 u0 L; {# d3 w* H1 B$ T/ ~% ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ @3 S# D: ?. R% ]2 V5 P
t) G/ O, e# yInduction Hypothesis: Let K >=2 be integers, support that
5 q, z) B3 o8 X5 K- d; T2 i+ |, J K^3 – K can by divided by 3.# I- N' o5 ?/ W" J2 T/ h$ i
7 }" j8 y, Y$ B) L# U8 H, ZNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 _/ v* M d4 U5 o" h
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( l( ~( x+ `* | EThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- K) [# z8 q: X" O6 S& I = K^3 + 3K^2 + 2K# s0 Y# Z3 Y! e3 ^ z* X2 _
= ( K^3 – K) + ( 3K^2 + 3K)
0 Y ^& h4 H: Z: t \9 y- s = ( K^3 – K) + 3 ( K^2 + K)
. c/ I: z0 M6 J% u1 W; n; ~7 Rby 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
5 O' s2 Z0 |3 T5 I( y7 }) R% mSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' s1 N5 K/ y* b5 [- W) G* v
= 3X + 3 ( K^2 + K)6 p8 k. i3 w0 {1 a! k
= 3(X+ K^2 + K) which can be divided by 3: V% s' F h& R8 `
" {8 m( Y3 i2 L! k6 BConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.0 [9 M! X, O' F9 P3 f) I4 n: f
! z( \7 G+ ^* u; q
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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