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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)- H4 m+ h3 ~9 m$ O
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Proof: 8 N! ? N/ e4 A3 i7 V8 a0 ~: S
Let n >1 be an integer % c; L4 ~5 T8 q% ?* V9 U
Basis: (n=2)( s; ]5 V- |1 u' D+ ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
2 R+ [, ?) g8 B1 k/ ?+ ?
4 J$ J( x/ o5 x6 {Induction Hypothesis: Let K >=2 be integers, support that: p9 n- r# P; f: b8 b ]4 v% { O
K^3 – K can by divided by 3.% W2 ]. y H3 b( _" j* j
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 c9 c5 N+ z$ m7 a! O% O# l! ?+ Xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem6 ^" l# L! w' {6 c7 ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& S4 N/ ^1 J4 E4 _% l
= K^3 + 3K^2 + 2K" W, _% }8 {0 T& h0 h1 q$ |6 |
= ( K^3 – K) + ( 3K^2 + 3K)
) g# n, e) n, V* x: X! I: B = ( K^3 – K) + 3 ( K^2 + K)
v0 Q# y' n$ A; Jby 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
; z# v5 r8 F" XSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 ~0 f9 o# ~( s! s! E P = 3X + 3 ( K^2 + K)
% A4 n& b- }4 X1 D' e" M = 3(X+ K^2 + K) which can be divided by 3
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& ~* Y, l! }" g F# n" G7 p4 K; bConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 L2 J. R w- C/ z+ s& L
# e, ?% a* s; B$ q1 w[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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