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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)( l( Y7 z- V! c) M3 K
9 h0 P Z* W! `Proof: , K9 ~4 o% s$ g8 n, r1 t
Let n >1 be an integer
8 P/ G- l! m4 m" z' Q# }Basis: (n=2)3 G5 L( o$ T- B
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
8 d" E: r/ |- q7 u. X7 p. J) ?% @, ?; z3 C* ]% I
Induction Hypothesis: Let K >=2 be integers, support that1 \6 S$ e0 G* t/ U; ~# G7 O/ ?- ?
K^3 – K can by divided by 3.5 ~6 ], o* @$ n. P/ M/ {
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 W6 k- b1 J, c0 a
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; N; F/ J' q) u: TThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
N: h$ Y/ |* Z8 s8 \ = K^3 + 3K^2 + 2K
1 B( m* w) O& q& G2 N* G = ( K^3 – K) + ( 3K^2 + 3K)) {& q i' ~- ^) L D& L9 L* p
= ( K^3 – K) + 3 ( K^2 + K)
% [+ ?; ~3 x4 z, q1 d' M0 V0 Jby 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( J5 M# D7 iSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 {8 ]2 w5 H6 e' v& R = 3X + 3 ( K^2 + K); z( Z7 q1 [8 A9 b2 S* s
= 3(X+ K^2 + K) which can be divided by 3& z9 ?0 x% v8 [. Q! z" Q# N
W2 K& @! P- M( G- _4 ~
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. r8 g* c$ {5 z
8 f l9 M4 i8 {) A/ f: a! [- o% ?3 Q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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