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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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( a* D. j; e/ a d! y) a( oProof: ; |3 j- c( V# {! B7 g
Let n >1 be an integer
5 T( o+ ^# A# b3 q+ B5 d+ W$ NBasis: (n=2)4 E0 I+ a. c' g, T* Z0 Y: t
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
1 I) R, k; G, q) ~% A" X' ?7 a( B& x- ^* H9 z, v
Induction Hypothesis: Let K >=2 be integers, support that& j9 X) I2 e9 w& Y
K^3 – K can by divided by 3.
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, w2 V7 M- X5 N! wNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 r8 P, z& y) z3 W; j) n
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
[% l3 A) r& [( r, f* @5 mThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
6 N S6 H6 q5 `9 C = K^3 + 3K^2 + 2K
% S2 J. n7 @2 _0 w0 q( W( \ = ( K^3 – K) + ( 3K^2 + 3K)+ _8 l+ U/ U2 {0 D! e w* H/ I
= ( K^3 – K) + 3 ( K^2 + K)
# c" f8 q2 ^& k E- lby 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 d' z D8 K" ^* r" I8 T
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)3 u3 u. t: r4 p* j: g
= 3X + 3 ( K^2 + K)
1 N! a, N# w4 O = 3(X+ K^2 + K) which can be divided by 3 `# G/ n Z- Q$ y
5 L% F4 k% k% ~. b& p, U4 i% tConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., ` U' i8 M" w5 a8 ?3 Z" {" M. ]3 H
0 S/ i3 |/ i0 d5 m* a$ i& J[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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