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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). A' ^. {5 y6 \8 K$ G6 y3 J
+ b4 u% H0 _- k N0 O/ L0 ?Proof: $ n# z2 U7 S" Q8 p- p8 S
Let n >1 be an integer 4 H5 Z3 ?- U- M8 z2 q. R
Basis: (n=2)! l, R" I: @" D/ o7 f
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: _1 z8 k# e" G+ a& ~" r$ I
+ ?$ a: a2 P. T# m) {
Induction Hypothesis: Let K >=2 be integers, support that; f7 ?1 a' _7 Y; A! n* N
K^3 – K can by divided by 3.6 @7 d5 w$ q6 l3 a* @2 { L! I
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( Q3 ?0 K# o. z( U3 E
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" L6 k: C7 j& S7 `8 J1 ?6 k
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ ^) }% X' w) e3 O! e. \0 i = K^3 + 3K^2 + 2K
' o- {+ O$ T6 \$ @* }0 P( R = ( K^3 – K) + ( 3K^2 + 3K)9 ^9 {6 r+ {- p) V8 ~ d- A
= ( K^3 – K) + 3 ( K^2 + K)( b0 s3 ^( @, t) S% g
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
% z2 r4 G2 s( ^+ G8 {So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ T6 r8 Q9 |2 C& I' T6 g8 T/ M5 w
= 3X + 3 ( K^2 + K)
' b7 b3 m/ C& Z" A8 D' B) u* k = 3(X+ K^2 + K) which can be divided by 3/ K' |( o9 ^* M- K, H! u
6 ]& V6 Q0 `9 X7 a! g' v
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% s" i9 k% ]0 n+ w
/ Z8 `& s! u+ k6 T% G
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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