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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: . |& H6 A3 x! s' {
Let n >1 be an integer
3 t% E2 u% y0 o9 o2 P& f, h) {Basis: (n=2)0 @# v" R( N+ ]+ y
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
( k/ [$ C, Q' P K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! ?4 C- j2 h; Qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem. b' v; S$ Q2 ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 f7 H2 c- _5 H/ U# i
= K^3 + 3K^2 + 2K7 C* `- {7 t: p
= ( K^3 – K) + ( 3K^2 + 3K)3 ~8 m8 r$ O% k2 t
= ( K^3 – K) + 3 ( K^2 + K)% }( J+ a9 \2 X1 t6 C& z; I
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
# D7 H4 J: F1 h0 i7 F$ v4 c1 }So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)6 S( h1 |& F& {' O
= 3X + 3 ( K^2 + K)
( Q5 F0 `7 M8 f" O( T8 x% k8 W = 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 >1.3 b1 E$ q) p2 s# T3 C) R$ U. \
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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