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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: 6 ~0 ^% b; _3 z
Let n >1 be an integer 6 y( R2 e& g3 i9 Q+ ~
Basis: (n=2)0 T% Z) ^8 m& _
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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: q0 |# U9 ?6 a" S+ n: TInduction Hypothesis: Let K >=2 be integers, support that
5 X2 I, r3 V% E K^3 – K can by divided by 3.# [' q9 S; k0 u! L
" t$ E* g$ B6 V1 [Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3$ e8 f3 W* d3 X. d% V
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
) W5 H& ?* x3 r Q3 e- n0 M; |9 @4 ^Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& B! r# {% `+ p) t; I
= K^3 + 3K^2 + 2K: T6 H. I5 Y! R+ C) A( I- r0 a
= ( K^3 – K) + ( 3K^2 + 3K)" r. j; R5 d& g; A6 C1 q$ a# y
= ( K^3 – K) + 3 ( K^2 + K)
9 ]1 e0 o# D( f7 Qby 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- q1 {5 K$ P( J6 A Z7 w6 X; p) V
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); D( G% h% g1 Q) p& M' t
= 3X + 3 ( K^2 + K)2 ^8 M6 {; {: Z3 W% k& u
= 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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8 H7 k# s4 T5 l6 L8 k$ ?4 ?! s% k[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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