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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)2 l- N# _0 p2 y* t. f
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Proof: 2 [! r6 N! z6 l9 p: h
Let n >1 be an integer ( V, C: v x, T! t' I \
Basis: (n=2)* v% H$ m! P+ V( c8 i. I0 `
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 39 F5 [- o; ^3 T; \, c4 r
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Induction Hypothesis: Let K >=2 be integers, support that* E7 h7 T) V# ?5 @" b5 K/ c
K^3 – K can by divided by 3.
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7 f, m! o3 p# s2 aNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. |6 X. a' x( _9 k0 hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
U5 S2 q7 B/ S2 @) [/ KThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
8 S4 r8 d. x4 H# }8 R = K^3 + 3K^2 + 2K
& m' b. z0 C0 P$ s = ( K^3 – K) + ( 3K^2 + 3K)! y9 Y, G" ^. _. \# B8 B6 p$ ?4 O/ \7 I
= ( K^3 – K) + 3 ( K^2 + K)+ ^7 a/ h/ b5 q, W
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
2 K% `) I0 d# {* kSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ h' l' P! C) ~& S0 ?4 g3 B7 }
= 3X + 3 ( K^2 + K)6 F4 `2 A* L$ Z; r
= 3(X+ K^2 + K) which can be divided by 3
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7 t0 R! a& x! pConclusion: 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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