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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)& i. T& U# Q; D' h) v5 ]0 F- ]* N, [
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Proof: 5 t, b3 p/ ]4 |& r$ ]3 \9 P
Let n >1 be an integer 7 m3 `' `6 Y8 E
Basis: (n=2) o4 z* d C! e" x* {2 S2 v
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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, f U- S: M* nInduction Hypothesis: Let K >=2 be integers, support that
9 _) ~3 i; g6 F K^3 – K can by divided by 3.; M% Q P! E- I
: @$ }/ t! w, N- }Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
~. j1 D7 N. g8 c& F7 i% M9 `8 G' {6 ysince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ _ |5 Z! j7 b$ o9 ?' u4 i$ z0 qThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ |% c" m0 u+ Q f = K^3 + 3K^2 + 2K
! W# @. a1 o7 l7 j2 ` = ( K^3 – K) + ( 3K^2 + 3K): O5 n7 {9 A7 J8 I: ?! a
= ( K^3 – K) + 3 ( K^2 + K)0 B+ K2 \6 F6 l9 v
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
6 Z! a: ]1 S4 {; ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)7 [5 C) f/ E t/ Q) D
= 3X + 3 ( K^2 + K). C5 Y V7 d3 I: t
= 3(X+ K^2 + K) which can be divided by 38 G" a l' {: l# N, h: R4 ]
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.# ^3 R) v* B- H4 p$ J& E
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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