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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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0 l8 J! c! @ \Proof: 9 V6 ~$ ?7 U h* V$ H; _
Let n >1 be an integer
" L D+ a5 h+ D5 {8 iBasis: (n=2)
& E D. W: m% D' H! U 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( e2 n+ K) j$ Y$ X9 a
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Induction Hypothesis: Let K >=2 be integers, support that. U, {, u! q% b/ ^7 I; }
K^3 – K can by divided by 3.. l7 q8 ^. ^, n/ @; L
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ d' f& {2 s% T( u
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ l* g; o5 P! B3 ^1 e- PThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)# |" Z8 W7 ?5 _9 R
= K^3 + 3K^2 + 2K
+ i/ i9 y {( Z = ( K^3 – K) + ( 3K^2 + 3K)
6 W \7 \+ ]7 v U) x, w' R$ I. F = ( K^3 – K) + 3 ( K^2 + K)- J& e0 B0 g+ H: e7 E5 {
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
+ M0 d, u5 T z) u8 ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 B8 L+ O* \! e4 W! A+ G' a = 3X + 3 ( K^2 + K)
3 k6 S+ m2 _* O) h" @% _8 B& @1 q& | = 3(X+ K^2 + K) which can be divided by 3& Z1 `1 K9 A+ P, I
# D0 c& X$ B9 a" d4 Z+ [, P: j9 _
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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