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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)) O% ~. j4 |: u6 u5 B: B3 f
, L8 Q5 P" }9 v3 [Proof:
* m$ a2 t) z1 p* a( vLet n >1 be an integer
) W+ b9 t: {; L' z3 ?, P# VBasis: (n=2)
1 i" y# f1 A/ F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
( P' q, Q& @: [7 V* f3 J' l) o# N0 W1 t) a, X+ H" A
Induction Hypothesis: Let K >=2 be integers, support that/ ^% y, ^4 m( t0 d( ?
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% [4 M/ c$ f5 dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem$ p, G: G3 |0 q$ g
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, @' @- i# {$ B; n = K^3 + 3K^2 + 2K
8 A O3 y- H& c6 B& ~2 t# t8 N# j = ( K^3 – K) + ( 3K^2 + 3K)
8 ~' ]& z+ D$ ^ = ( K^3 – K) + 3 ( K^2 + K)# t$ @' A: o" B7 ]
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>08 _& x' J. u1 Y
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K): c$ W& ~9 n1 N+ s% r
= 3X + 3 ( K^2 + K)
: ]" V2 j: O# c: P1 z. d = 3(X+ K^2 + K) which can be divided by 3
" p: b1 g) r- G3 W- B# H: t8 _' I( g; r; R \% e) [2 N3 {
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ o& W! r7 ?2 q* m2 o4 H1 O- O
- N; I$ p& t4 ?. y. \[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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