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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)) U2 C8 Z" r7 n( q2 i; a
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Proof: + v2 ^7 j2 G8 U2 ~( O+ [4 \. F. q
Let n >1 be an integer 3 q5 X) R/ Y3 o5 k8 U( z6 W" S( A# B
Basis: (n=2)
/ |* Y- M2 e# D 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
1 `+ D( p* V3 A- L5 N+ }1 s' A+ H5 R p4 P2 Q
Induction Hypothesis: Let K >=2 be integers, support that X. A" g; {; K8 C
K^3 – K can by divided by 3.9 O9 V+ |: S4 G' y0 g2 E
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
1 o) W5 _2 }! P( D, v/ B) Ysince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 o* d$ Y) ^1 E7 B' V
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* i* h5 C% f1 d* h9 h = K^3 + 3K^2 + 2K
! c/ q" S. u9 ?# [- ^ = ( K^3 – K) + ( 3K^2 + 3K)8 w2 c% J; P7 y8 I, Z" M* k
= ( K^3 – K) + 3 ( K^2 + K); `# k( n. {% ^2 W
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>04 R9 ~% P) y: x6 ]
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' ?( F1 ]6 ]. d$ U = 3X + 3 ( K^2 + K)
/ U3 o( u6 e) C H$ _' \" W- `# E = 3(X+ K^2 + K) which can be divided by 3/ q2 ^: ^0 [7 u
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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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# N4 _* i7 \7 F8 {& a[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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