 鲜花( 53)  鸡蛋( 0)
|
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)
. I7 G$ H5 l3 n1 n% C K( r: ]
& [) ]3 Q" B2 R# HProof:
( }2 Y: j1 p6 V$ w3 g( a& n' x* C, mLet n >1 be an integer : }$ ~3 i" c- ?. a" E7 X
Basis: (n=2)
$ A3 e0 E. n+ U$ p6 K3 ~ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
2 t9 Q# M; h( ^* g4 p0 x+ {) s& p' a" k% }8 d
Induction Hypothesis: Let K >=2 be integers, support that
( r- V9 Z7 X- N+ M* w& X" _2 I K^3 – K can by divided by 3.
7 p" _0 Z. C+ s4 X) v% v* d& T6 @) b" E; B6 \* X
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, w6 v" y. q$ F: Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' N3 g% b8 W9 w1 O' c; C4 I$ i% aThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)- l |; L% H. o7 c/ ]" v
= K^3 + 3K^2 + 2K: v$ W+ m" H7 Y
= ( K^3 – K) + ( 3K^2 + 3K)
( P p- Q F5 y = ( K^3 – K) + 3 ( K^2 + K)
) F& G+ A! n! D/ b' w5 P0 D$ \0 B+ x4 fby 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. V/ |5 _; D, e$ o4 B
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 V4 ?* v, H- A6 K2 m
= 3X + 3 ( K^2 + K)3 ~. }9 ?: K4 }+ E
= 3(X+ K^2 + K) which can be divided by 34 `1 e6 [. ~" t% o% z( F
6 \; @2 J/ `* D. D% c! z
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
8 L7 q7 S+ |. d2 W- g' T
& \+ D# I1 s' q# m[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|