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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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2 S& `$ w _* M( X/ f) w4 J$ P6 LProof:
, Y0 O' g1 Q1 {; G# h( K xLet n >1 be an integer 4 G O( i% q" A' ^: h
Basis: (n=2)( W* E4 g0 X4 d
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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* o( I' ?( T7 a4 s% LInduction Hypothesis: Let K >=2 be integers, support that: x/ s4 d$ x% o1 b1 ]( D% S& |
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ f% S! f3 \" }6 @8 r" Bsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 V8 Y1 E1 D! c+ I! G. u- w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. p6 X) ]$ `* S( t! Z = K^3 + 3K^2 + 2K
2 ]& V! P2 l+ {; ]5 @; z1 c = ( K^3 – K) + ( 3K^2 + 3K)+ Y5 P7 y7 `+ f
= ( K^3 – K) + 3 ( K^2 + K)) t5 _8 X6 Y6 t' {4 ^" K
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- X6 |2 i$ X) l$ k ~
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 k( }) s$ h7 Z+ w; H = 3X + 3 ( K^2 + K)* F; I" f$ ]+ L$ F& O) p0 M
= 3(X+ K^2 + K) which can be divided by 3
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+ ~( U& u [9 K$ _' VConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.0 Q( O6 B" F- r1 E9 j4 a0 `7 b& A/ ^
3 F# u4 {4 x2 h4 O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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