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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)% W6 A9 u' P1 y: ?# x# r4 E
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Proof:
! Y/ \& \: b5 H( VLet n >1 be an integer
0 X$ K/ Y+ u3 T( |; B3 ]Basis: (n=2)
x n: d1 g. S 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) Y4 t/ Y0 K8 M' H: F! g6 X
1 _$ _ M0 B, ]& O# ]: n8 z% m# wInduction Hypothesis: Let K >=2 be integers, support that
9 b. K: Y \* y" U, `$ \ K^3 – K can by divided by 3.
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3 g/ c# Z+ H. M6 ~1 O nNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 37 @5 [" d: Z; o' L5 v6 z0 m! M1 q
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
5 i. N* X3 I& @7 `% h2 }Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, [' f6 @/ C7 b# J6 a = K^3 + 3K^2 + 2K
5 q* b7 K/ m, c = ( K^3 – K) + ( 3K^2 + 3K)
* c. W* Y7 L( \ = ( K^3 – K) + 3 ( K^2 + K)6 b* ~& {5 Z9 U0 K5 A
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; O: x* [4 ^' r7 z( ?/ i
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: @1 i$ T, R0 w5 ?) H. [( } = 3X + 3 ( K^2 + K)% N4 _6 p; r v* o8 p4 ?" q' p
= 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.! h4 ?& x$ e- N- _: x8 r+ F: d$ q% e9 Z
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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