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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)$ F$ t5 F& |5 T2 i. d: K* `. r
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Proof: " q; R3 k5 c- x$ r4 \6 G
Let n >1 be an integer
: D& ^8 q- _" F) @6 V \; [Basis: (n=2)
& N3 L) A' { [8 i( ]) ^3 S6 _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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4 f$ T- ^8 H0 h* L1 N4 d* zInduction Hypothesis: Let K >=2 be integers, support that( p1 X( \' W7 [+ Z
K^3 – K can by divided by 3.. M# E Z. Q+ t2 v9 T
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& E. ]% e5 W4 |: l5 ?- T. x, f5 _" Qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 ^! R8 G! Y- U1 b! [Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); p: N$ i! I `0 X1 Q* x
= K^3 + 3K^2 + 2K7 t1 Q4 \) z8 A, O
= ( K^3 – K) + ( 3K^2 + 3K)2 n# ]" a- e1 y) p# l' e/ N+ I
= ( K^3 – K) + 3 ( K^2 + K)7 F* C& j }: o7 L
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 ]5 |( {/ @: ?+ Y
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' c2 ?( A4 u' k3 _7 Z+ P = 3X + 3 ( K^2 + K)0 l: |" N6 ]/ ]9 R
= 3(X+ K^2 + K) which can be divided by 3
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2 m6 g* T$ R- W1 I- c. i5 p# k- VConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; x6 X0 U7 f6 D# k# v
' Z0 S6 d/ A. C& P7 w# e: F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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