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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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Proof: & T S$ J m( b4 q6 u4 h
Let n >1 be an integer
, ?. g, H7 \, D3 uBasis: (n=2)
I6 j( T. q( l1 K* ?+ t5 m 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: W' d. f* R! I, t4 f" q5 z
4 X/ y- r# A! P0 fInduction Hypothesis: Let K >=2 be integers, support that
0 q/ R" t$ L4 G4 a K^3 – K can by divided by 3." L- \- V/ S5 r' |3 C- H
7 m1 g; s- v* t# p- }: HNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 p2 ~( H. K2 \) D1 k1 i4 q7 Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, [ H" q/ I+ u5 s' x# P
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ |7 ?1 w& Y: b' A P0 ?8 r+ T
= K^3 + 3K^2 + 2K$ z( h0 t8 y4 D
= ( K^3 – K) + ( 3K^2 + 3K)
+ ?% o. @8 \) I = ( K^3 – K) + 3 ( K^2 + K)0 `+ X( F b5 `" X# R$ w1 s' X7 ~
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* C/ u' X: {1 _9 ~5 H* v0 P1 O
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ H" W" ^3 I) b- W( `% I = 3X + 3 ( K^2 + K)
" P& \3 q: j* c' q- x = 3(X+ K^2 + K) which can be divided by 3+ W& _% E/ Y+ q% k- U8 G% n
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; w! G$ U/ t. L# {+ M
7 l+ U- @& `8 p$ \. M. a e* H[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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