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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)9 k0 f/ a0 k( F6 U. |( s+ a1 ]
( G! z) t% k9 v" ?Proof: 8 ]% ^' b6 ^8 A
Let n >1 be an integer G1 ] u v2 z k
Basis: (n=2)8 h9 U4 F( \6 h6 X# Q3 B; A: G
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
" X# r: r1 I8 E7 z
6 f" [* \+ r. A2 e# S7 q6 [Induction Hypothesis: Let K >=2 be integers, support that
, ?1 ]5 d/ `% `& Z3 Z4 Y+ r K^3 – K can by divided by 3.
4 K1 u# Y7 _! O8 r4 A3 N
+ P C) B! f6 o4 e- U( eNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' p: b- q% `- K. m% m# f8 O) z# usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( _) S! k+ B+ W8 r5 DThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ [4 B: m; Z6 L1 T = K^3 + 3K^2 + 2K
- L% x. T4 R& l- | = ( K^3 – K) + ( 3K^2 + 3K)
/ S2 b# c" w! c/ g8 Q+ D/ B6 C = ( K^3 – K) + 3 ( K^2 + K)5 ~: ~ h1 @$ k9 _# @1 ~
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; K9 J( O2 W3 A1 H4 k; f7 q0 I$ p# g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# D) V0 I, C( T! d- c) B = 3X + 3 ( K^2 + K)1 [+ w5 ]! o" U8 v5 s8 D* J9 Y
= 3(X+ K^2 + K) which can be divided by 3 }8 R& B. V5 O8 i! e7 d& H
) ^, j4 ]% k7 |: rConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 y# W$ F- p' d, V' r/ _
$ K* k" `, h) o1 |' j[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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