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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: / e- D& I! r" k. ?* P) ~& W
Let n >1 be an integer ; f# ^# e0 T! D8 F+ x* s0 y
Basis: (n=2)+ }9 e& ?# j. x1 c& A4 Z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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# k. ?- l& E" X/ w' N% j8 S' G2 ]Induction Hypothesis: Let K >=2 be integers, support that: o2 [6 a+ K( a0 A; G3 M/ B
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 32 g9 m( Y( _7 i: R
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( @3 ~; I5 Y6 O4 F- j" p
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 n" ^7 A& Y$ e5 N+ t- l" F
= K^3 + 3K^2 + 2K
$ b3 z4 Y6 @4 ?& S( q9 S" I$ M1 ?0 } = ( K^3 – K) + ( 3K^2 + 3K)
7 G% A" V2 v b4 j& U5 N = ( K^3 – K) + 3 ( K^2 + K)
( {1 o- b0 {5 v% F1 `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. h0 y! Z, k/ |3 L" u+ S1 |9 J
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( y1 \9 {( a2 U4 l3 e9 O: v/ g = 3X + 3 ( K^2 + K)7 Y. n7 p% v- c. s& U5 D
= 3(X+ K^2 + K) which can be divided by 3* B" T9 [9 I- p- l6 s
3 A4 R+ N/ U8 I! S oConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.' C) }$ I% D! Y* u$ Z
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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