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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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/ f" j; \! ~5 f0 C; v6 ^8 ]Proof: " E8 z ~, k2 L( k
Let n >1 be an integer
8 q4 f7 ^6 i% `) O) K( D. H2 N+ \Basis: (n=2)
7 T; Q/ x- n, b' p3 e @# _9 ? 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3* J* i+ ~$ W B, i2 Z0 b7 W1 _
2 o- {9 R- n9 T6 b* l; aInduction Hypothesis: Let K >=2 be integers, support that
7 I; d# m1 {: j! I. ^ K^3 – K can by divided by 3.
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( [" J# P* X: |, F7 L* V- QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ z" o# V' t3 i7 a$ l
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' }2 n5 r* Z1 W1 }! J; I& p8 i: q. {
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): g6 J. i; T0 r' E) F7 V+ J6 l1 }+ n
= K^3 + 3K^2 + 2K7 O% U" i7 e8 @( }$ S; f
= ( K^3 – K) + ( 3K^2 + 3K)! H% S& T t- X1 L
= ( K^3 – K) + 3 ( K^2 + K)
; R o0 j3 e) \7 Nby 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+ p( s$ d0 j i1 B. i) z8 d
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% _% Z4 ^# b$ s+ U# S8 B8 ^
= 3X + 3 ( K^2 + K)% @. O" x2 o% y( g9 l2 @4 u
= 3(X+ K^2 + K) which can be divided by 3
& J2 o/ Q* a# C/ [* \7 c. M- _3 P" \) \2 b3 ] U
Conclusion: 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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