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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)$ W# a6 O" U2 m: b9 y
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Proof: & W3 o2 Y" y/ v1 N: R* r3 g9 t
Let n >1 be an integer
8 H: P5 B' j2 z7 A, {Basis: (n=2)
+ l4 E3 \' x- u- n5 E$ ]+ r8 z; g. R 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% \, U8 c1 V# T; j! S. ]& R, o
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Induction Hypothesis: Let K >=2 be integers, support that
L2 e- ? r% P0 ] K^3 – K can by divided by 3.6 R9 I g+ {& t0 f7 H0 l# S
. @2 ~, u* O: S/ B" aNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' F& i4 t0 g2 w$ D! o
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem) v6 s4 R- ~8 n4 U4 {( p- V
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ u# u$ B9 ]6 J) g3 H = K^3 + 3K^2 + 2K5 F' |' \: k6 f; i/ j9 n
= ( K^3 – K) + ( 3K^2 + 3K); N6 Q2 f% p; O
= ( K^3 – K) + 3 ( K^2 + K)
. q* A( e4 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>0
) a4 g3 M9 X( Q9 JSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. l/ L% ? m9 [: ~! B = 3X + 3 ( K^2 + K)* I i3 \+ q* t$ e3 x
= 3(X+ K^2 + K) which can be divided by 3, E7 V3 h; b, S" D' i6 a6 d
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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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- O- D: p. c) L: t ~' u[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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