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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:
8 ?$ [3 \% T3 g5 ~6 `Let n >1 be an integer
' y+ I9 e( J5 W2 N. IBasis: (n=2)1 W W3 S' ], ?3 s4 m. Q0 m4 p6 T
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 j1 @5 f2 C. o9 y7 {1 u' H' ~
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Induction Hypothesis: Let K >=2 be integers, support that& n; B0 v/ E' }
K^3 – K can by divided by 3.
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/ A; N/ Q6 A2 n: y9 ^" y* aNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 A& M1 s0 b- \& j1 M# isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 T/ [! @) E0 Z. ?8 s
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 v9 Y( e6 K) t: b a+ x& _ = K^3 + 3K^2 + 2K/ G7 c' D; k ]0 w g) p! e
= ( K^3 – K) + ( 3K^2 + 3K)4 ~9 Y' G" q% z5 L" s# E0 Z0 Y3 V
= ( K^3 – K) + 3 ( K^2 + K)
, q- I1 q- A, {2 O7 V: n2 pby 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/ ^" f( X: N' i0 S8 I* {
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) A& K& U/ u4 J. H3 t = 3X + 3 ( K^2 + K): y/ o* I' j) m/ Q; I2 G$ C' p
= 3(X+ K^2 + K) which can be divided by 3) V: B$ r* L3 N& C4 I; C
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 A1 y' V, p& m# Z. P/ y: d/ w
: `1 J: X- L" r[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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